Question

Giải phương trình:

\(\sqrt{5x^2+27x+25}-5\sqrt{x+1}=\sqrt{x^2-4}\)

P/S:e đăng hộ cho 1 bạn nha.Ths mn

Answer 1

ĐK:\(\hept{\begin{cases}5x^2+27x+25\ge0\\x+1\ge0\\x^2-4\ge0\end{cases}}\)(*)

\(pt\Leftrightarrow\sqrt{5x^2+27x+25}=5\sqrt{x+1}+\sqrt{x^2-4}\)

\(\Leftrightarrow5x^2+27x+25=25x+25+x^2-4+10\sqrt{\left(x+1\right)\left(x^2-4\right)}\)

\(\Leftrightarrow4x^2+2x+4=10\sqrt{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow2x^2+x+2=5\sqrt{\left(x^2-x-2\right)\left(x+2\right)}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2-x-2}=a\\\sqrt{x+2}=b\end{cases}}\)\(\Rightarrow2a^2+3b^2=5ab\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\2a=3b\end{cases}}\)..............