Đáp án:
`b)S=\{2;3\}`
`c)S=\{0\}`
Giải thích các bước giải:
`b)(x-2)(2x-1)=5(x-2)`
`⇔(x-2)(2x-1)-5(x-2)=0`
`⇔(x-2)[(2x-1)-5]=0`
`⇔(x-2)(2x-1-5)=0`
`⇔(x-2)(2x-6)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\2x-6=0 \end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\2x=6\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy `S=\{2;3\}`
`c)ĐKXĐ:x\ne 5;x\ne -5`
`(x+5)/(x-5)-(x-5)/(x+5)=(x(x+25))/(x^2-25)`
`⇔(x+5)^2/((x-5)(x+5))-(x-5)^2/((x+5)(x-5))=(x(x+25))/((x-5)(x+5))`
`⇒(x+5)^2-(x-5)^2=x(x+25)`
`⇔(x+5+x-5)(x+5-x+5)=x^2+25x`
`⇔2x.10=x^2+25x`
`⇔20x=x^2+25x`
`⇔x^2+25x-20x=0`
`⇔x^2+5x=0`
`⇔x(x+5)=0`
\(⇔\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0(TM)\\x=-5(KTM)\end{array} \right.\)
Vậy `S=\{0\}`
