Đáp án:
`a)``S=\{1+\sqrt 19;1-\sqrt 19\}`
`b)S=∅`
Giải thích các bước giải:
`a)ĐKXĐ:x\ne -9`
`\frac{x^2}{x+9}=2`
`⇔\frac{x^2}{x+9}=\frac{2(x+9)}{x+9}`
`⇒x^2=2(x+9)`
`⇔x^2=2x+18`
`⇔x^2-2x-18=0`
`⇔x^2-2x+1-19=0`
`⇔(x-1)^2-(\sqrt 19)^2=0`
`⇔(x-1-\sqrt 19)(x-1+\sqrt 19)=0`
\(⇔\left[ \begin{array}{l}x-1-\sqrt 19=0\\x-1+\sqrt 19=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1+\sqrt 19(tm)\\x=1-\sqrt 19(tm)\end{array} \right.\)
Vậy `S=\{1+\sqrt 19;1-\sqrt 19\}`
`b)ĐKXĐ:x\ne -9`
`\frac{x^2}{x+9}=-20`
`⇔\frac{x^2}{x+9}=\frac{-20(x+9)}{x+9}`
`⇒x^2=-20x-180`
`⇔x^2+20x+180=0`
`⇔x^2+20x+100+80=0`
`⇔(x+10)^2+80=0`
Vì `(x+10)^2≥0∀x`
`⇒(x+10)^2+80\ne 0`
Vậy `S=∅`
