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Đáp án+Giải thích các bước giải:

`a)3x+1=\frac{-7}{2}`

`<=>\frac{2(3x+1)}{2}=\frac{-7}{2}`

`<=>2(3x+1)=-7`

`<=>6x+2=-7`

`<=>6x=-7-2`

`<=>6x=-9`

`<=>x=-9/6`

`<=>x=-3/2`

Vậy `S={-3/2}`

$\\$

`b)(3x-5)^2-(9x^2-25)=0`

`<=>(3x-5).(3x-5)-(3x-5)(3x+5)=0`

`<=>(3x-5)[(3x-5)-(3x+5)]=0`

`<=>(3x-5)(3x-5-3x-5)=0`

`<=>-10(3x-5)=0`

`<=>`\(\left[ \begin{array}{l}3x-5=0\\-10=0(loại)\end{array} \right.\) 

`<=>3x-5=0`

`<=>3x=0+5`

`<=>3x=5`

`<=>x=5/3`

Vậy `S={5/3}`

$\\$

`c)\frac{x-4}{5}+\frac{3x+2}{10}=7`

`<=>\frac{2(x-4)}{10}+\frac{3x+2}{10}=70/10`

`<=>2(x-4)+3x+2=70`

`<=>2x-8+3x+2=70`

`<=>5x-6=70`

`<=>5x=70+6`

`<=>5x=76`

`<=>x=76/5`

Vậy `S={76/5}`

$\\$

`d)\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{2}{x^2-4}+1(đk:x\ne-2; x\ne2)`

`<=>\frac{(x+1)(x+2)}{(x-2)(x+2)}-\frac{5(x-2)}{(x+2)(x-2)}=\frac{2}{(x-2)(x+2)}+\frac{(x+2)(x-2)}{(x+2)(x-2)}`

`=>(x+1)(x+2)-5(x-2)=2+(x+2)(x-2)`

`<=>x^2+2x+x+2-5x+10=2+x^2-4`

`<=>x^2-2x+12=x^2-2`

`<=>x^2-x^2-2x=-2-12`

`<=>-2x=-14`

`<=>x=(-14)/-2`

`<=>x=7(tm)`

Vậy `S={7}`