Question

\(\dfrac{\sqrt{21+x}+\sqrt{21-x}}{\sqrt{21+x}-\sqrt{21-x}}=\dfrac{21}{x}\)

Answer 1

ĐKXĐ: -21\(\le x\le\)21

Đặt \(\left\{{}\begin{matrix}\sqrt{21+x}=a\\\sqrt{21-x}=b\end{matrix}\right.\left(a,b\ge0\right)\) (a\(\ne\)b)

Ta có \(\left\{{}\begin{matrix}21+x=a^2\\21-x=b^2\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}a^2+b^2=42\\a^2-b^2=2x\end{matrix}\right.\)

Pt đã cho trở thành \(\dfrac{a+b}{a-b}=\dfrac{a^2+b^2}{a^2-b^2}\)

<=> \(\left(a+b\right)^2\)(a-b)=(\(a^2+b^2\))(a-b)

<=> (a-b)2ab=0

\(\text{​​}\text{​​}\left[{}\begin{matrix}a=b\left(loai\right)\\a=0\left(tm\right)\\b=0\left(tm\right)\end{matrix}\right.\)

Thay vào ta tìm dc S=\(\left\{21,-21\right\}\)