Với `x \ne +-2` có:

`M=[x^3]/[x^2-4]-x/[x-2]-2/[x+2]`

`M=[x^3-x(x+2)-2(x-2)]/[(x-2)(x+2)]`

`M=[x^3-x^2-2x-2x+4]/[(x-2)(x+2)]`

`M=[x^3-x^2-4x+4]/[(x-2)(x+2)]`

`M=[x^2(x-1)-4(x-1)]/[x^2-4]`

`M=[(x-1)(x^2-4)]/[x^2-4]`

`M=x-1`

\(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

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

Với  có:

a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5

b: \(M=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{2x-5}{x\left(x+5\right)}\)

\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)

A = 3x^3 +6x^2 + 3xy^3

x= 1 phần 2 ;  p = -1 phần 3

A=3.1 phần 2^3 . -1 phần 3     + 6.(1 phần 2)^2 . (-1 Phần 3)^2+3 1 phần 2 . (-1 phần 3)^3

=-1 phần 8      + -1 phần 2 - 1 phần 2

= -1 phần 4

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

b: \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

c: Để M=-1/2 thì 2(x+1)=-2

=>x+1=-1

hay x=-2(nhận)

a, ĐKXĐ của M là :

\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ne0\\2\left(1-x\right)\left(1+x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

b, \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

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

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

c, Để giá trị của biểu thức M bằng \(\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2x-2}=\dfrac{1}{2}\)

\(\Leftrightarrow2x-2=2\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\)

Vậy x = 2 thfi giá trị của M bằng 1/2

a.

M xác định \(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)

b.

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

c.

M=1/2

\(\Leftrightarrow\dfrac{1}{2x+2}=\dfrac{1}{2}\\ \Leftrightarrow2x+2=2\\ \Leftrightarrow x=0\)

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

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

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

c: Để M=1/2 thì 2x+2=2

hay x=0(nhận)

a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)

b: Thay x=11 vào M, ta được:

\(M=\dfrac{8}{11-3}=1\)

a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)

\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)

b) Thay \(x=11\left(TM\right)\) vào biểu thức M: 

\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)

M=\(\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

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

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

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

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

Để biểu thức M có nghĩa thì

\(2\left(x+1\right)\ne0\)

=>\(x+1\ne0\)

\(x\ne-1\)

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

Mà M=\(\dfrac{1}{2}\)

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

=>2=2(x+1)

2=2x+2

x=0

Vậy x=0

a/ Để m có nghĩa thì:

\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ne0\\2\left(1-x^2\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\x^2\ne1\end{matrix}\right.\)\(\Leftrightarrow x\ne\pm1\)

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

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

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

c/ M = \(-\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{1}{2\left(x+1\right)}=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{-1}{2}\cdot2\left(x+1\right)=1\Leftrightarrow-x-1=1\Leftrightarrow-x=2\Leftrightarrow x=-2\)