Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{Đk:}x\ne-\frac{1}{2}\Rightarrow P=\frac{4x^2\left(x+2\right)-\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(4x^2-1\right)\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(2x-1\right)\left(x+2\right)}{2x+1}\)
\(=\frac{2x^2+4x-x-2}{2x+1}=\frac{3}{2}\Rightarrow2x^2+3x-2=3x+\frac{3}{2}\Leftrightarrow2x^2-\frac{7}{2}=0......\)
\(P\text{ nguyên }\Rightarrow2x^2+3x-2⋮2x+1\Leftrightarrow2x^2+3x-2-\left(x+1\right)\left(2x+1\right)⋮2x+1\Leftrightarrow-3⋮2x+1....\)
a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)
ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)
b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)
\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)
c) \(P=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{\left(x+2\right)\left(2x+1-2\right)}{2x+1}=\frac{\left(x+2\right)\left(2x+1\right)-2\left(x+2\right)}{2x+1}\)
\(=x+2-\frac{2x+2}{2x+1}=x+2-1-\frac{1}{2x+1}\)
để P nguyền khi zà chỉ khi
\(1⋮2x+1\)
\(=>2x+1\inƯ\left(1\right)=\pm1\)
=>\(\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
a) đk: \(x\ne-\frac{1}{2}\)
b) \(P=\frac{3}{2}\Leftrightarrow\frac{4x^3+4x^2-x-2}{4x^2+4x+1}=\frac{3}{2}\)
\(\Leftrightarrow8x^3+8x^2-2x-4=12x^2+12x+3\)
\(\Leftrightarrow8x^3-4x^2-14x-7=0\)
Cardano ra
c) \(P=\frac{4x^3+4x^2-x-2}{4x^2+4x+1}=x-\frac{2x+2}{4x^2+4x+1}\)
Xét delta tìm khoảng giá trị của biến P
BÀI 1:
a) \(ĐKXĐ:\) \(x-3\)\(\ne\)\(0\)
\(\Leftrightarrow\)\(x\)\(\ne\)\(3\)
b) \(A=\frac{x^3-3x^2+4x-1}{x-3}\)
\(=\frac{\left(x^3-3x^2\right)+\left(4x-12\right)+11}{x-3}\)
\(=\frac{x^2\left(x-3\right)+4\left(x-3\right)+11}{x-3}\)
\(=x^2+4+\frac{11}{x-3}\)
Để \(A\)có giá trị nguyên thì \(\frac{11}{x-3}\)có giá trị nguyên
hay \(x-3\)\(\notinƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta lập bảng sau
\(x-3\) \(-11\) \(-1\) \(1\) \(11\)
\(x\) \(-8\) \(2\) \(4\) \(14\)
Vậy....
a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)
Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)
Q = \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Q = \(\frac{x^2-2x-x+2+4x+4-3x-6}{\left(x+2\right)\left(x-2\right)}\)
Q = \(\frac{x^2-2x}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)
b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3
Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)
=> x + 2 = 3(x - 3)
=> x + 2 = 3x - 9
=> x - 3x = -9 - 2
=> -2x = -11
=> x = 11/2 (tm)
Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)
c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3
Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)
Để M \(\in\)Z <=> 3 \(⋮\)x - 3
=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng:
| x - 3 | 1 | -1 | 3 | -3 |
| x | 4 | 2 (ktm) | 6 | 0 |
Vậy ...
1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x+2\right)\left(x-2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)
\(\left(x+2\right)\ne0\Rightarrow x\ne-2\)
\(\left(x-2\right)\ne0\Rightarrow x\ne2\)
Vậy để biểu thức xác định thì : \(x\ne\pm2\)
b) để C=0 thì ....
1, c , bn Nguyễn Hữu Triết chưa lm xong
ta có : \(/x-5/=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)
thay x = 7 vào biểu thứcC
\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...
thay x = 3 vào C
\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)
=> ko tìm đc giá trị C tại x = 3
a, ĐKXĐ: x\(\ne\) 1;-1;2
b, A= \(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{2x^2-2x}{2\left(x+1\right)\left(x-1\right)}+\frac{2x+2}{2\left(x+1\right)\left(x-1\right)}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-2}{x+1}\)
=\(\frac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{x-2}{x-1}\)
c, Khi x= -1
→A= \(\frac{-1-2}{-1-1}\)
= -3
Vậy khi x= -1 thì A= -3
Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^
a,ĐKXĐ:x#1; x#-1; x#2
b,Ta có:
A=\(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{x\left(x-1\right)2}{\left(x+1\right)\left(x-1\right)2}+\frac{\left(x+1\right)2}{\left(x-1\right)\left(x+1\right)2}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{x-2}\)
=\(\frac{2x^2-2x+2x+2+4x}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{x-2}{x+1}\)
c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả
d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên
\(\Leftrightarrow x-2⋮x+1\)
\(\Leftrightarrow x+1-3⋮x+1\)
Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)
Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên
a) ĐKXĐ của phương trình : \(4x^2+4x+1\ne0\)\(\Rightarrow x\ne-\frac{1}{2}\)
b) \(P=\frac{4x^3+8x^2-x-2}{4x^2+4x+1}\)
\(\Rightarrow P=\frac{\left(4x^3-x\right)+\left(8x^2-2\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P=\frac{x\left(4x^2-1\right)+2\left(4x^2-1\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{3}{2}\)\(\Rightarrow P\left(x\right)=2\left(x+2\right)\left(2x-1\right)=3\left(2x+1\right)\)
\(\Rightarrow P\left(x\right)=4x^2+6x-6-\left(6x+3\right)=0\)
\(\Rightarrow P\left(x\right)=4x^2-9=0\)\(\Rightarrow P\left(x\right)=x^2=\frac{9}{4}\)
\(\Rightarrow P\left(x\right)=x^2=\sqrt{\frac{9}{4}}\)\(\Rightarrow P\left(x\right)=\frac{3}{2}\)
câu c) cx tương tự
a, x khác -1/2
b, x=\(\frac{\sqrt{7}}{2}\)