Đáp án:
`a)S=\{-1\}`
`b)S=\{0\}`
`c)S=\{-4\}`
Giải thích các bước giải:
`a)ĐKXĐ:x\ne 3;x\ne -3`
`(x-2)/(x+3)-(x+2)/(x-3)=(x^2+1)/(x^2-9)-1`
`<=>\frac{(x-2)(x-3)-(x+2)(x+3)}{(x-3)(x+3)}=(x^2+1)/(x^2-9)-(x^2-9)/(x^2-9)`
`=>(x-2)(x-3)-(x+2)(x+3)=x^2+1-(x^2-9)`
`<=>(x^2-3x-2x+6)-(x^2+3x+2x+6)=x^2+1-x^2+9`
`<=>(x^2-5x+6)-(x^2+5x+6)=10`
`<=>x^2-5x+6-x^2-5x-6=10`
`<=>-10x=10`
`<=>x=-1(TM)`
Vậy `S=\{-1\}`
`b)ĐKXĐ:x\ne 2;x\ne 5`
`(3x)/(x-2)-x/(x-5)-(2x^2)/(x^2-7x+10)=0`
`<=>(3x)/(x-2)-x/(x-5)-(2x^2)/(x^2-2x-5x+10)=0`
`<=>(3x)/(x-2)-x/(x-5)-(2x^2)/(x(x-2)-5(x-2))=0`
`<=>\frac{3x(x-5)-x(x-2)-2x^2}{(x-2)(x-5)}=0`
`=>3x(x-5)-x(x-2)-2x^2=0`
`<=>3x^2-15x-x^2+2x-2x^2=0`
`<=>-13x=0`
`<=>x=0(TM)`
Vậy `S=\{0\}`
`c)ĐKXĐ:x\ne 3;x\ne -3;x\ne -7/2`
`13/((x-3)(2x+7))+1/(2x+7)=6/(x^2-9`
`<=>\frac{13(x+3)+(x-3)(x+3)}{(2x+7)(x-3)(x+3)}=\frac{6(2x+7)}{(x-3)(x+3)(2x+7)}`
`=>13(x+3)+(x-3)(x+3)=6(2x+7)`
`<=>13x+39+x^2-9=12x+42`
`<=>x^2+13x+30=12x+42`
`<=>x^2+13x+30-12x-42=0`
`<=>x^2+x-12=0`
`<=>x^2-3x+4x-12=0`
`<=>x(x-3)+4(x-3)=0`
`<=>(x-3)(x+4)=0`
`<=>x-3=0` `\text{hoặc}` `x+4=0`
`<=>x=3(KTM)` `\text{hoặc}` `x=-4(TM)`
Vậy `S=\{-4\}`
