Question

help me phần b plssssssssssss

Answer 1

`a)6/(x-5)+2/(x-8)=18/((x-5)(8-x))-1` `(ĐK:x<>5;x<>8)`

`<=>(6(x-8))/((x-5)(x-8))+(2(x-5))/((x-5)(x-8))=-18/((x-5)(x-8))-((x-5)(x-8))/((x-5)(x-8))`

`<=>6(x-8)+2(x-5)=-18-((x-5)(x-8))`

`<=>6x-48+2x-10=-18-x^2+13x-40`

`<=>8x-58=-x^2+13x-58`

`<=>x^2-13x+8x=0`

`<=>x^2-5x=0` 

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

`<=>x=0(tm)` hoặc `x=5(ktm)`

`b)(x^2-x)/(x+3)-x^2/(x-3)=(7x^2-3x)/(9-x^2)` `(ĐK:x<>+-3)` 

`<=>(x^2-x)/(x+3)-x^2/(x-3)=(3x-7x^2)/(x^2-9)`

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

`<=>x^3-x^2-3x^2+3x-x^3-3x^2=3x-7x^2`

`<=>-7x^2=3x-7x^2`

`<=>3x=0`

`<=>x=0(tm)` 

Answer 2

`6/{x-5}+2/{x-8}=18/{(x-5)(8-x)}-1`

`(ĐKXĐ:x ne 5;8)`

`=>6(x-8)+2(x-5)=-18-(x-5)(x-8)`

`<=>6x-49+2x-10=-18-(x^2-13x+40)`

`<=>8x-40=-x^2+13x-40`

`<=>x^2-5x=0`

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

`<=>x=0` hoặc `x=5`

Thử lại ta thấy `x=0` thỏa mãn

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