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 = \(\left\{\frac{11}{27}\right\}\)

<=> 

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

\(B=\frac{\left|2x-1\right|+x}{3x^2-22x+7}=\frac{1-2x+x}{3x^2-21x-x+7}=\frac{1-x}{\left(x-7\right)\left(3x-1\right)}\)

Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12

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

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

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