Đáp án + Giải thích các bước giải:
`a//(2x-5)/(x+5)=3` `(ĐKXĐ:x\ne-5)`
`⇔3(x+5)=2x-5`
`⇔3x+15=2x-5`
`⇔3x-2x=-15-5`
`⇔x=-20(TM)`
Vậy `S={-20}`
`b//(x^{2}-6)/(x)=x+(3)/(2)` `(ĐKXĐ:x\ne0)`
`⇔(2(x^{2}-6))/(2x)=(2x^{2})/(2x)+(3x)/(2x)`
`⇒2(x^{2}-6)=2x^{2}+3x`
`⇔2x^{2}-12=2x^{2}+3x`
`⇔2x^{2}-2x^{2}-3x=12`
`⇔-3x=12`
`⇔x=-4(TM)`
Vậy `S={-4}`
`c//((x^{2}+2x)-(3x+6))/(x-3)=0` `(ĐKXĐ:x\ne3)`
`⇔(x^{2}+2x)-(3x+6)=0`
`⇔x(x+2)-3(x+2)=0`
`⇔(x+2)(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=-2(TM)\\x=3(KTM)\end{array} \right.\)
Vậy `S={-2}`
`d//(5)/(3x+2)=2x-1` `(ĐKXĐ:x\ne-(2)/(3))`
`⇔(2x-1)(3x+2)=5`
`⇔6x^{2}+x-2-5=0`
`⇔6x^{2}+x-7=0`
`⇔(6x^{2}-6x)+(7x-7)=0`
`⇔6x(x-1)+7(x-1)=0`
`⇔(x-1)(6x+7)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\6x+7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1(TM)\\x=-\frac{7}{6}(TM)\end{array} \right.\)
Vậy `S={1;-(7)/(6)}`
