1.  A = -4 phần x+2

2.  2x^2 + x = 0 => x = 0 hoặc x = -1/2

    Với x = 0 thì A = -2

    Với x = -1/2 thì A = -8/3

3.   A = 1/2 =>  -4 phần x + 2  = 1/2

                  <=> -8 = x + 2 

                   <=> x = -10

4.   A nguyên dương => A > 0

                               => -4 phần x + 2 > 0

      Do -4 < 0 nên -4 phần x + 2 > 0 khi x + 2 < 0

                                                        => x < -2

A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x 
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x 
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x 
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x) 
= - x+6/x+2

\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)

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

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

Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)

a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)

Quy đồng :

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

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

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

b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)

\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)

c ) MTC : \(\left(x+2\right)^3\)

\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)

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

\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)

bài này là giải phương trình hả bn ?

1.

<=> 7 - 2x - 4 = -x - 4

<=> -2x + x = -4 -7 + 4

<=> -x = -7

<=> x = 7

       Vậy S = { 7 }

2.

<=> \(\frac{2\left(3x-1\right)}{6}\)= \(\frac{3\left(2-x\right)}{6}\)

<=> 2( 3x - 1 ) = 3( 2 - x )

<=> 6x -2 = 6 - 3x

<=> 6x + 3x = 6 + 2

<=> 9x = 8

<=> x = \(\frac{8}{9}\)

       Vậy S =  \(\left\{\frac{8}{9}\right\}\)

3.

<=> \(\frac{6x+10}{3}-\frac{x}{2}=5-\frac{3x+3}{4}\)

<=> \(\frac{4\left(6x+10\right)}{12}-\frac{6x}{12}=\frac{60}{12}-\frac{3\left(3x+3\right)}{12}\)

<=> 4( 6x + 10 ) - 6x = 60 - 3( 3x + 3 )

<=> 24x + 40 - 6x = 60 - 9x -9

<=> 18x + 40 = 51 - 9x

<=> 18x + 9x = 51 - 40

<=> 27x = 11

<=> x = \(\frac{11}{27}\)

       Vậy S = \(\left\{\frac{11}{27}\right\}\)

<=> 

\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x=a\)

=> \(\left(a-6\right)\left(a+6\right)=a^2-36\ge-36\)

\(x\left(x+5\right)=0\) thì biểu thức nhỏ nhất

<=> x = 0 hoặc x = -5