Gửi đến bạn Vũ:\(x-\frac{6}{7}=5 \frac{2}{3}.x\)\(\Rightarrow x=5 \frac{2}{3}.x \frac{6}{7}\)\(\Rightarrow x=\frac{41}{7} \frac{2}{3}x\)\(\Rightarrow0=\frac{41}{7} \frac{2}{3}x-x\)\(\Rightarrow0=\frac...

PT

Phương Trình Hai Ẩn

12 tháng 6 2017 - olm

Gửi đến bạn Vũ:\(x-\frac{6}{7}=5+\frac{2}{3}.x\)\(\Rightarrow x=5+\frac{2}{3}.x+\frac{6}{7}\)\(\Rightarrow x=\frac{41}{7}+\frac{2}{3}x\)\(\Rightarrow0=\frac{41}{7}+\frac{2}{3}x-x\)\(\Rightarrow0=\frac{41}{7}+x\left(-\frac{1}{3}\right)\)\(\Rightarrow-\frac{41}{7}=x.\left(-\frac{1}{3}\right)\)\(\Rightarrow...

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#Toán lớp 7

3

DC

Đỗ Công Tùng

12 tháng 6 2017

gửi đến bạn vũ thì gửi cho bạn đấy chứ đăng lên đây làm gì.

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K

kaitokid2017

12 tháng 6 2017

đỗ công tùng đúng đó đăng làm j

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LN

lu nguyễn

3 tháng 4 2020

tìm các giới hạn sau:

a, \(\lim\limits_{x\rightarrow-3}\frac{x+\sqrt{3-2x}}{x^2+3x}\)

b, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+9}+\sqrt{x+16}-7}{x}\)

c, \(\lim\limits_{x\rightarrow\frac{1}{2}}\frac{8x^2-1}{6x^2-5x+1}\)

d, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}\)

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

3 tháng 4 2020

Vậy nó ko phải dạng vô định, cứ thay số trực tiếp

\(=\frac{2}{0}=+\infty\)

Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy

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NV

Nguyễn Việt Lâm

Giáo viên

3 tháng 4 2020

Sry mình ko nhớ 1 chữ về vật lý luôn :<

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NA

Ngọc Ánh Nguyễn Thị

11 tháng 3 2020

1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\) 2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\) 3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\) 4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\) 5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\) 6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\) 7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\) 8,...

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#Toán lớp 11

9

AH

Akai Haruma

Giáo viên

12 tháng 3 2020

Bài 2:

\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)

Bài 3:

\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)

\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)

Bài 4:

\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)

Bài 5:

\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)

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AH

Akai Haruma

Giáo viên

12 tháng 3 2020

Bài 6:

\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)

Bài 7:

\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)

Bài 8:

\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)

Bài 9:

\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)

\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)

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TP

Trần Phương Thảo

15 tháng 3 2020

Bài 1 a. \(\lim\limits_{x\rightarrow+\infty}\frac{1+2\sqrt{x}-x}{x+3}\) b. \(\lim\limits_{x\rightarrow+\infty}\frac{x^3+3x-1}{x^2\sqrt{x}+x}\) c. \(\lim\limits_{x\rightarrow-\infty}\frac{x+2\sqrt{1-x}}{1-x}\) Bài 2: Tính các giới hạn sau biết \(\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1\) a. \(\lim\limits_{x\rightarrow0}\frac{1-\cos x}{1-\cos3x}\) b. \(\lim\limits_{x\rightarrow0}\frac{\cot x-\sin x}{x^3}\) c. \(\lim\limits_{x\rightarrow\infty}\frac{x.\sin...

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#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 1:

\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)

Bài 2:

\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)

Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)

Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

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TP

Trần Phương Thảo

15 tháng 3 2020

a. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}\) f. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7-3}}{2-\sqrt{x+3}}\) b. \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}\) g. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\) c. \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}\) h. \(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}\) d. \(\lim\limits_{x\rightarrow1}\frac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}\) k....

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#Toán lớp 11

1

MC

meo con

15 tháng 3 2020

a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)

b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)

c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)

d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)

e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)

f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)

g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)

h)

\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)

k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

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TP

Trần Phương Thảo

15 tháng 3 2020

Bài 1 a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}\) b. \(\lim\limits_{x\rightarrow1}\frac{x^6-5x^5+x}{\left(1-x\right)^2}\) c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\) d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+x\right)}{x^5+x^2}\) Bài 2 a. \(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}\) b. \(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)\) c....

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#Toán lớp 11

2

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 1:

\(a=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)

\(b=\frac{1-5+1}{0}=\frac{-3}{0}=-\infty\)

\(c=\lim\limits_{x\rightarrow1}\frac{x\left(1+2x\right)\left(1+3x\right)+2x\left(1+3x\right)+3x}{x}=\lim\limits_{x\rightarrow1}\left[\left(1+2x\right)\left(1+3x\right)+2\left(1+3x\right)+3\right]=1+2+3=6\)

\(d=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-1}{5x^4+2x}=\frac{4}{0}=+\infty\)

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NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

Bài 2:

\(a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)

\(c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n\)

\(\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)\)

\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)\)

\(=...\)

\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}\)

\(=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]\)

\(=1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

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LN

lu nguyễn

14 tháng 3 2020

tìm các giới hạn sau:

a, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x^2}-1}{x}\)

b,\(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)

c, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\)

d, \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{4x-2}}{x-2}\)

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 3 2020

\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)

\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)

\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)

\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)

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TP

Trần Phương Thảo

1 tháng 4 2020

Bài 1 a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}\) b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}\) c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\) d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}\) Bài 2 a. \(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}\) b. \(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)\) ...

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#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

3 tháng 4 2020

Bài 1:

a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)

b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)

c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)

d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)

\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)

Bài 2:

\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)

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LN

lu nguyễn

14 tháng 3 2020

tìm các giới hạn sau:

a, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\)

b, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+x^2}-1}{x^2}\)

c, \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}+\sqrt{x+7}-5}{x-2}\)

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 3 2020

\(a=\frac{0-1}{0-1}=1\)

\(b=\lim\limits_{x\rightarrow0}\frac{\frac{x^2}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}}{x^2}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}=\frac{1}{3}\)

\(c=\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}-2+\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{x+7}+3}}{x-2}=\lim\limits_{x\rightarrow2}\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{x+7}+3}\right)\)

\(=\frac{1}{\sqrt{4}+2}+\frac{1}{\sqrt{9}+3}=\frac{5}{12}\)

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LN

lu nguyễn

4 tháng 4 2020

tìm các giới hạn sau: a; \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\) b, \(\lim\limits_{x\rightarrow+\infty}\frac{\left(2x-3\right)^2\left(4x+7\right)^3}{\left(3x^3+1\right)\left(10x^2+9\right)}\) c,\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\) ( bài này k hiểu mk tính kiểu gì 1 cái ra \(+\infty\) một cái ra \(-\infty\)) d, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}.\sqrt{1+6x}-1}{x}\) e,...

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#Toán lớp 11

2

NV

Nguyễn Việt Lâm

Giáo viên

5 tháng 4 2020

\(a=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{9+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{9+\sqrt{x+7}}}{x-1}=\frac{29}{36}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(2-\frac{3}{x}\right)^2.x^3\left(4+\frac{7}{x}\right)^3}{x^3\left(3+\frac{1}{x^3}\right).x^2\left(10+\frac{9}{x^2}\right)}=\frac{2.4}{3.10}=\frac{4}{15}\)

\(c=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\frac{\frac{-4x^2}{\sqrt{1+4x}+2x+1}+\frac{8x^3+12x^2}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\frac{-4}{\sqrt{1+4x}+2x+1}+\frac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=\frac{-4}{1+1}+\frac{12}{1+1+1}=2\)

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NV

Nguyễn Việt Lâm

Giáo viên

5 tháng 4 2020

\(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}\left(\sqrt{1+4x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{4x\sqrt{1+6x}}{x\left(\sqrt{1+4x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{6x}{x\left(\sqrt{1+6x}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{4\sqrt{1+6x}}{\sqrt{1+4x}+1}+\lim\limits_{x\rightarrow0}\frac{6}{\sqrt{1+6x}+1}=\frac{4}{1+1}+\frac{6}{1+1}=5\)

\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{2x\sqrt[3]{1+4x}}{x\left(\sqrt{1+2x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{4x}{x\left(\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{2\sqrt[3]{1+4x}}{\sqrt{1+2x}+1}+\lim\limits_{x\rightarrow0}\frac{4}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}=\frac{2}{1+1}+\frac{4}{1+1+1}=\frac{7}{3}\)

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