Đáp án:

Giải thích các bước giải:

`1)` `3/(x+3)-1/(x-2)=5/(2x+6)` 

`ĐKXĐ : x ne -3;2`

`<=> (6(x-2))/(2(x+3)(x-2))-(2(x+3))/(2(x+3)(x-2))=(5(x-2))/(2(x+3)(x-2))`

`<=> (6(x-2)-2(x+3))/(2(x+3)(x-2))=(5(x-2))/(2(x+3)(x-2))`

`=> 6(x-2)-2(x+3)=5(x-2)`

`<=> 6x-12-2x-6=5x-10`

`<=> 6x-2x-5x=-10+6+12`

`<=> -x=8`

`<=> x=-8 (TM)`

`-> S={-8}`

`2) `

`1/(2x-2)+2/(x-2)=5/(3x+3)`

`ĐKXĐ : x ne 1;2;-1`

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

`<=> (3(x-2)(x+1)+12(x+1)(x-1))/(6(x-2)(x+1)(x-1))=(10(x-1)(x-2))/(6(x-2)(x+1)(x-1))`

`=> 3(x-2)(x+1)+12(x+1)(x-1)=10(x-1)(x-2)`

`<=> 3(x^2-x-2)+12(x^2-1)=10(x^2-3x+2)`

`<=> 3x^2-3x-6+12x^2-12=10x^2-30x+20`

`<=> 3x^2+12x^2-10x^2-3x+30x=20+12+6`

`<=> 5x^2+27x=38`

`<=> 5x^2+27x-38=0`

`<=> 5(x^2+27/5x-38/5)=0`

`<=> x^2+27/5x-38/5=0`

`<=> x^2+2.x. 27/10+(27/10)^2-(27/10)^2-38/5=0`

`<=> (x+27/10)^2-1489/100=0`

`<=> (x+27/10)^2=1489/100`

`<=>` \(\left[ \begin{array}{l}x+\dfrac{27}{10}=\dfrac{\sqrt{1489}}{10}\\x+\dfrac{27}{10}=-\dfrac{\sqrt{1489}}{10}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{1489}-27}{10}\\x=\dfrac{\sqrt{-1489}-27}{10}\end{array} \right.\) 

`-> S={\frac{\sqrt{-1489}-27}{10};\frac{\sqrt{1489}-27}{10}}`

Đáp án + Giải thích các bước giải:

`1//(3)/(x+3)-(1)/(x-2)=(5)/(2x+6)` `(ĐKXĐ:x\ne{-3;2})`

`⇔(6(x-2))/(2(x+3)(x-2))-(2(x+3))/(2(x+3)(x-2))=(5(x-2))/(2(x+3)(x-2))`

`⇔6(x-2)-2(x+3)=5(x-2)`

`⇔6x-12-2x-6=5x-10`

`⇔6x-2x-5x=12+6-10`

`⇔-x=8`

`⇔x=-8(TM)`

Vậy `S={-8}`

`2//(1)/(2x-2)+(2)/(x-2)=(5)/(3x+3)` `(ĐKXĐ:x\ne{±1;2})`

`⇔(1)/(2(x-1))+(2)/(x-2)=(5)/(3(x+1))`

`⇔(3(x+1)(x-2))/(6(x-1)(x+1)(x-2))+(12(x+1)(x-1))/(6(x-2)(x+1)(x-1))=(10(x-1)(x-2))/(6(x+1)(x-1)(x-2))`

`⇒3(x+1)(x-2)+12(x+1)(x-1)=10(x-1)(x-2)`

`⇔3(x^{2}-x-2)+12(x^{2}-1)=10(x^{2}-3x+2)`

`⇔3x^{2}-3x-6+12x^{2}-12=10x^{2}-30x+20`

`⇔(3x^{2}+12x^{2}-10x^{2})+(-3x+30x)-(6+12+20)=0`

`⇔5x^{2}+27x-38=0`

`⇔x^{2}+(27)/(5)x-(38)/(5)=0`

`⇔x^{2}+2.x.(27)/(10)+((27)/(10))^{2}-(1489)/(100)=0`

`⇔(x+(27)/(10))^{2}=(1489)/(100)`

`⇔x+(27)/(10)=±(\sqrt{1489})/(10)`

`⇔` \(\left[ \begin{array}{l}x=\frac{-27+\sqrt{1489}}{10}(TM)\\x=\frac{-27-\sqrt{1489}}{10}(TM)\end{array} \right.\) 

Vậy `S={\frac{-27+\sqrt{1489}}{10};\frac{-27-\sqrt{1489}}{10}}`