Question

\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}=4\)

giải phương trình bằng cách đặt ẩn phụ

Answer 1

Đặt \(t=\sqrt{x+2}+\sqrt{5-x}\Rightarrow t^2=7+2\sqrt{\left(x+2\right)\left(5-x\right)}\)

=> \(\sqrt{\left(x+2\right)\left(5-x\right)}=\dfrac{t^2-7}{2}\); t² \(\ge\)7

=> t + \(\dfrac{t^2-7}{2}=4\) <=> \(\dfrac{t^2+2t-15}{2}=0\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-5\end{matrix}\right.\)

t = 3 <=> \(\sqrt{x+2}+\sqrt{5-x}=3\Rightarrow x+2+5-x+2\sqrt{\left(x+2\right)\left(5-x\right)=9}\)<=> \(\sqrt{\left(x+2\right)\left(5-x\right)}=1\Leftrightarrow\left(x+2\right)\left(5-x\right)=1\Leftrightarrow-x^2+3x+9=0\Leftrightarrow\left[{}\begin{matrix}\dfrac{3+3\sqrt{5}}{2}\\\dfrac{3-3\sqrt{5}}{2}\end{matrix}\right.\)