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

chết mk nhìn nhầm phần c bài 2 :

\(2,\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

Để P xác định 

\(\Rightarrow2-x\ne0\Rightarrow x\ne2\)

\(2+x\ne0\Rightarrow x\ne-2\)

\(x^2-4\ne0\Rightarrow x\ne0\)

\(x^2-3x\ne0\Rightarrow x\ne3\)

b, \(P=\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2+x\right)\left(2-x\right)}+\frac{2-x}{2+x}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(P=\left[\frac{4+4x+x^2}{\left(2-x\right)\left(2+x\right)}-\frac{4x^2}{\left(2+x\right)\left(2-x\right)}-\frac{4-4x+x^2}{\left(2+x\right)\left(2-x\right)}\right].\frac{x\left(2-x\right)}{x-3}\)

\(P=\left[\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\right].\frac{x\left(2-x\right)}{x-3}=\frac{4x\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

\(P=\frac{4x^2\left(2-x\right)}{\left(x-3\right)\left(2+x\right)}\)

d, ĐỂ \(p=\frac{8x^2-4x^3}{x^2-x-6}< 0\)

\(TH1:8x^2-4x^3< 0\)

\(\Rightarrow8x^2< 4x^3\)

\(\Rightarrow2< x\Rightarrow x>2\)

\(TH2:x^2-x-6< 0\Rightarrow x^2< x+6\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

c: Thay x=-2 vào A, ta được:

\(A=\dfrac{-2-1}{-2+1}=\dfrac{-3}{-1}=3\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(P=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)

a: ĐKXĐ: 

b: 

a) \(P=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)

a) \(ĐKXĐ:\) x khác + 3

\(b,P=\dfrac{3\left(x-3\right)+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(P=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(P=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

\(P=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(P=\dfrac{4}{x-3}\)

c) \(P=4=\dfrac{4}{x-3}=4=x-3=1=x=4\)

a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

b: \(P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)

c: Để P=4 thì x-3=1

hay x=4

\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4}{x-3}\\ c,P=4\Leftrightarrow\dfrac{4}{x-3}=4\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

a) Giá trị của biểu thức A đã co xác định 

\(\Leftrightarrow\hept{\begin{cases}x^2+x\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x+1\right)\ne0\\x\ne-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}}\)

Vậy với \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)thì giá trị của biểu thức A đã cho được xác định .

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

b)

+) \(A=\left(\frac{1}{x^2+x}+\frac{1}{x+1}\right).x^2\)

\(A=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{x+1}\right).x^2\)

\(A=\frac{1+x}{x\left(x+1\right)}.x^2\)

\(A=\frac{1}{x}.x^2=x\)

+) 

Ta có :

\(A\left(x^2-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

<=> x = 0 ( không thỏa mãn ĐKXĐ) hoặc x = 1( thỏa mãn ĐKXĐ) hoặc x = -1 ( Không thỏa mãn ĐKXĐ)

Vậy với x = 1 thì \(A\left(x^2-1\right)=0\)

\(a.ĐKXĐ:\hept{\begin{cases}x^2+x\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x+1\right)\ne0\\x\ne-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0vax\ne-1\\x\ne-1\end{cases}\Leftrightarrow}x\ne0vax\ne-1}\)

\(A=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{x+1}\right).x^2\)

\(=\frac{1+1x}{x\left(x+1\right)}.x^2\)

\(=\frac{1+1x}{x^2+x}.x^2\)

\(=\frac{1+1x}{x}\) với \(x\ne0\)và \(x\ne-1\)

\(a,ĐK:x\ne0;x\ne1;x\ne\pm2\\ b,A=\left[\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}\right]\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(x+2\right)}\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{4x\left(x+1\right)\cdot x}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x^2}{x+2}\)

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

\(\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\)

viết lại biểu thức 

a) \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}=\dfrac{\left(x-1-2\right)\left(x-1+2\right)}{\left(2x+1-x-2\right)\left(2x+1+x+2\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{3\left(x-1\right)\left(x+1\right)}\)  (1)

\(\Rightarrow\) ĐKXĐ: \(x\ne\pm1\)

b) \(\left(1\right)=\dfrac{x-3}{3x-3}\) (2)

c) Thay \(x=-3;x=1\) vào (2) ta có: \(\left\{{}\begin{matrix}B=\dfrac{-3-3}{3.\left(-3\right)-3}=\dfrac{1}{2}\\B=\dfrac{1-3}{3.1-3}=0\end{matrix}\right.\)

d) \(B=5\Rightarrow\dfrac{x-3}{3x-3}=5\Leftrightarrow x-3=15x-15\Leftrightarrow x=\dfrac{6}{7}\)