đề đây \(\sqrt{x}+\sqrt{5-x}=x^2-5x+7\)
\(\Leftrightarrow-x^2+5x+\sqrt{x}+\sqrt{5-x}-7=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\x=4\end{cases}}\)
$\Leftrightarrow-x^2+5x+\sqrt{x}+\sqrt{5-x}-7=0$⇔−x2+5x+√x+√5−x−7=0
\(\Leftrightarrow\hept{\begin{cases}x=1\\x=4\end{cases}}\)
ĐK:........
\(pt\Leftrightarrow\left(x+2-3\sqrt{x}\right)+\left(7-x-3\sqrt{5-x}\right)+3\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\frac{\left(x+2\right)^2-9x}{x+2+3\sqrt{x}}+\frac{\left(7-x\right)^2-9\left(5-x\right)}{7-x+3\sqrt{5-x}}+3\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\frac{1}{x+2+3\sqrt{x}}+\frac{1}{7-x+\sqrt{5-x}}+2\right)=0\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow x\in\left\{1;\text{ }4\right\}\)