Question

Giải phương trình: $(\sqrt{x+1}-2\sqrt{4-x})\sqrt{2x^2+18}-5x+15=0$

Answer 1

\(\left(\sqrt{x+1}-2\sqrt{4-x}\right)\sqrt{2x^2+18}-5x+15=0\)

ĐKXĐ \(\left\{{}\begin{matrix}x+1\ge0\\4-x\ge0\\2x^2+18\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le4}\)

\(\Leftrightarrow\dfrac{x+1-4\left(4-x\right)}{\sqrt{x+1}+2\sqrt{4-x}}\sqrt{2x^2+18}-5x+15=0\)

\(\Leftrightarrow\dfrac{5x-15}{\sqrt{x+1}+2\sqrt{4-x}}\sqrt{2x^2+18}-\left(5x-15\right)=0\)

\(\Leftrightarrow\left(5x-15\right)\left(\dfrac{\sqrt{2x^2+18}}{\sqrt{x+1}+2\sqrt{4-x}}-1\right)=0\)

\(\Leftrightarrow5x-15=0\Leftrightarrow x=3\)

Vậy x=3