Câu E vs K thôi ạ H em giải r ạ

Giải thích các bước giải:

`e)`

`(2x-14)/(x(x-7))-2/x=(2x)/(x-7)`

`ĐKXĐ:` `{(x\ne0),(x-7\ne0):}⇔{(x\ne0),(x\ne7):}`

`⇔` `(2x-14)/(x(x-7))-(2(x-7))/(x(x-7))=((2x)x)/(x(x-7))`

`⇔` `2x-14-(2x-14)=2x^2`

`⇔` `2x-14-2x+14=2x^2`

`⇔` `-2x^2=0`

`⇔` `x^2=0(KTM)`

Vậy `S=\emptyset`

`k)`

`⇔` `(x-5)/(x+5)-(x+5)/(x-5)=(x^2-21)/(x^2-25)`

`ĐKXĐ:` `{(x+5\ne0),(x-5\ne0):}⇔{(x\ne-5),(x\ne5):}`

`⇔` `((x-5)(x-5))/((x+5)(x-5))-((x+5)(x+5))/((x+5)(x-5))=(x^2-21)/((x+5)(x-5))`

`⇔` `x^2-10x+25-(x^2+10x+25)=x^2-21`

`⇔` `x^2-10x+25-x^2-10x-25=x^2-21`

`⇔` `-20x=x^2-21`

`⇔` `-20x-x^2=-21`

`⇔` `-20x-x^2+21=0`

`⇔` `-20x-x^2+20+1=0`

`⇔` `(-20x+20)+(-x^2+1)=0`

`⇔` `20(1-x)+(1-x)(1+x)=0`

`⇔` `(1-x)(20-1-x)=0`

$\left[\begin{matrix} 1-x=0\\ 19-x=0\end{matrix}\right.$ `⇔` $\left[\begin{matrix} x=1(TM)\\ x=19(TM)\end{matrix}\right.$

Vậy `S={1;19}`