`c)` `(3x-2).(4x+5)=0`

`<=>`\(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}3x=2\\4x=-5\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=-\frac{5}{4}\end{array} \right.\)

Vậy `S={2/3;-5/4}`

`e)` `(3x-1).(2x-5)=(3x-1).(x+2)`

`<=>(3x-1).(2x-5)-(3x-1).(x+2)=0`

`<=>(3x-1).(2x-5-x-2)=0`

`<=>(3x-1).(x-7)=0`

`<=>`\(\left[ \begin{array}{l}3x-1=0\\x-7=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}3x=1\\x=7\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=\frac{1}{3}\\x=7\end{array} \right.\)

Vậy `S= {1/3;7}`

`g)` `x.(2x-9)=3x.(x-5)`

`<=>2x^2-9x=3x^2-15x`

`<=>2x^2-9x-3x^2+15x=0`

`<=>(2x^2-3x^2)+(-9x+15x)=0`

`<=>-x^2+6x=0`

`<=>-(x^2-6x)=0`

`<=>x^2-6x=0`

`<=>x.(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)

Vậy `S= {0,6}`

`f)` `(2x-3)^2=(2x-3).(x+1)`

`<=>(2x-3).(2x-3)-(2x-3).(x+1)=0`

`<=>(2x-3).(2x-3-x-1)=0`

`<=>(2x-3).(x-4)=0`

`<=>`\(\left[ \begin{array}{l}2x-3=0\\x-4=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}2x=3\\x=4\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=\frac{3}{2}\\x=4\end{array} \right.\)

Vậy `S={3/2,4}`

`h)` `3x-15=2x.(x-5)`

`<=>3.(x-5)=2x.(x-5)`

`<=>3.(x-5)-2x.(x-5)=0`

`<=>(3-2x).(x-5)=0`

`<=>` \(\left[ \begin{array}{l}3-2x=0\\x-5=0\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}2x=3\\x=5\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=5\end{array} \right.\)

Vậy `S={3/2,5}`

`s)` `(x+5)/(x-5)-(x-5)/(x+5)=20/(x^2-25)`             ĐKXĐ : `x` $\neq$ `+-5`

`<=>(x+5)/(x-5)-(x-5)/(x+5)=20/((x-5).(x+5))`  

`<=>(x+5)^2/((x-5).(x+5))-(x-5)^2/((x-5).(x+5))=20/((x-5).(x+5))`  

`=>(x+5)^2-(x-5)^2=20`

`<=>(x+5-x+5).(x+5+x-5)=20`

`<=>10.2x=20`

`<=>20x=20`

`<=>x=1`  ( thỏa mãn ĐKXĐ )

Vậy `S={1}`

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

`<=>1/(x-1)+2/(x+1)=x/((x-1).(x+1))` 

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

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

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

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

`<=>2x=1`

`<=>x=1/2` ( thỏa mãn ĐKXĐ )

Vậy `S={1/2}`

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