`a)`

`5-(x+6)=4.(3-2x)`

`<=>-x-1=12-8x`

`<=>-x+8x=12+1`

`<=>7x=13`

`<=>x=13/7`

Vậy `S={13/7}`

`b)` `3-4x.(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>(8x^2-8x^2)+(-100x-x)=-300-3`

`<=>-101x=-303`

`<=>x=3`

Vậy `S={3}`

`c)` `(5x+2)/6-(8x-1)/3=(4x+2)/5-5`

`<=>(5(5x+2))/30-(10(8x-1))/30=(6(4x+2))/30-150/30`

`<=>5.(5x+2)-10.(8x-1)=6.(4x+2)-150`

`<=>25x+10-80x+10=24x+12-150`

`<=>25x-80x-24x=12-150-10-10`

`<=>-79x=-158`

`<=>x=2`

Vậy `S={2}`

`d)` `(x+2)/6+x=(2x-3)/8`

`<=>(4.(x+2))/24+(24x)/24=(3.(2x-3))/24`

`<=>4.(x+2)+24x=3.(2x-3)`

`<=>4x+8+24x=6x-9`

`<=>4x+24x-6x=-9-8`

`<=>22x=-17`

`<=>x=-17/22`

Vậy `S={-17/22}`

`e)` `(x-1)/x+1/(x+1)=(2x-1)/(x^2+x)`                    ĐKXĐ : `x` $\neq$ `0` và `x` $\neq$ `-1`

`<=>(x-1)/x+1/(x+1)=(2x-1)/(x.(x+1)` 

`<=>((x-1).(x+1))/(x.(x+1)+x/(x.(x+1)=(2x-1)/(x.(x+1))`

`<=>(x-1).(x+1)+x=2x-1`

`<=>x^2-1+x=2x-1`

`<=>x^2+x-2x=-1+1`

`<=>x^2-x=0`

`<=>x.(x-1)=0`

`<=>` \(\left[ \begin{array}{l}x=0(ktm)\\x-1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0(ktm)\\x=1\end{array} \right.\) 

Vậy `S={1}`

Đáp án + Giải thích các bước giải:

NL: chữ đẹp z -))
a) $5 - (x + 6) = 4(3 - 2x)$

$\Leftrightarrow 5 - x - 6 = 12 - 8x$

$\Leftrightarrow 5 - x - 6 - 12 + 8x = 0$

$\Leftrightarrow 7x - 13 = 0$

$\Leftrightarrow x = \dfrac{13}{7}$

Vậy $S = $`{13/7}`

b) $3 - 4x(25 - 2x) = 8x^2 + x - 300$

$\Leftrightarrow 3 - 100x + 8x^2 = 8x^2 + x - 300$

$\Leftrightarrow 8x^2 - 100x + 3 - 8x^2 - x + 300 = 0$

$\Leftrightarrow -101x + 303 = 0$

$\Leftrightarrow x = 3$

Vậy $S = $`{3}`

c) $\dfrac{5x +2}{6} - \dfrac{8x - 1}{3} = \dfrac{4x + 2}{5} - 5$

$\Leftrightarrow \dfrac{5(5x + 2) - 10(8x - 1)}{30} = \dfrac{6(4x + 2) - 150}{30}$

$\Leftrightarrow \dfrac{25x + 10 - 80x + 10}{30} = \dfrac{24x + 12 - 150}{30}$

$\Rightarrow -55x + 20 = 24x - 138$

$\Leftrightarrow -55x + 20 - 24x + 138 = 0$

$\Leftrightarrow -79x + 158 = 0$

$\Leftrightarrow x = 2$

Vậy $S = $`{2}`

d) $\dfrac{x + 2}{6} + x = \dfrac{2x - 3}{8}$

$\Leftrightarrow \dfrac{4(x + 2) + 24x}{24} = \dfrac{3(2x - 3)}{24}$

$\Rightarrow 4x +8 + 24x = 6x - 9$

$\Leftrightarrow 28x + 8 = 6x - 9$

$\Leftrightarrow 28x + 8 - 6x + 9 = 0$

$\Leftrightarrow 22x + 17 = 0$

$\Leftrightarrow x = \dfrac{-17}{22}$

 e) $\dfrac{x - 1}{x} + \dfrac{1}{x + 1} = \dfrac{2x - 1}{x^2 + x}(x \ne 0; -1)$

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

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

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

$\Rightarrow x^2 + x - 1 = 2x - 1$

$\Leftrightarrow x^2 + x - 1 - 2x + 1 =0$

$\Leftrightarrow x^2 - x = 0$

$\Leftrightarrow x(x - 1) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=0(l)\\x=1(n)\end{array} \right.\) 

Vậy $S = $`{1}`