\(N=\left(\dfrac{1-a^3}{1-a}-a\right):\dfrac{1-a^2}{1-a+a^3-a^2}\)
a) Rút gọn N
b) a bằng mấy để N < 0
c) Tìm a thuộc Z để N thuộc Z
K
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14 tháng 3 2022
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
6 tháng 4 2017
a) Để A và n thuộc Z => n+1 chia hết cho n-2
A=(n-2+3) chia hết cho n-2
=> 3 chia hết cho n-2
lập bảng=> n thuộc {3,1,5,9,(-1)}
b) A lớn nhất khi n-2 nhỏ nhất=> n-2=1
=> n=3
Nhớ tk cho mk nha!
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 8 2021
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)
c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)
\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)
\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)
12 tháng 8 2021
a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Thay x=36 vào A, ta được:
\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)
c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)
\(\Leftrightarrow4\sqrt{x}=2\)
hay \(x=\dfrac{1}{4}\)
a: \(N=\left(\dfrac{\left(1-a\right)\left(a^2+a+1\right)}{1-a}-a\right)\cdot\dfrac{a^3-a^2-a+1}{-\left(a^2-1\right)}\)
\(=\left(a^2+1\right)\cdot\dfrac{a^2\left(a-1\right)-\left(a-1\right)}{-\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\dfrac{\left(a-1\right)\left(a^2-1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\left(a-1\right)\)
b: Để N<0 thì \(-\left(a^2+1\right)\left(a-1\right)< 0\)
\(\Leftrightarrow\left(a^2+1\right)\left(a-1\right)>0\)
=>a-1>0
hay a>1