Question

Giải phương trình :

\(\dfrac{2}{\sqrt{x+1}+\sqrt{3-x}}=1+\sqrt{3+2x-x^2}\)

Answer 1

\(đặt:\sqrt{x+1}+\sqrt{3-x}=t\left(2\le t\le2\sqrt[]{2}\right)\)

\(pt\Leftrightarrow\dfrac{2}{t}=1+\sqrt{\dfrac{t^2-4}{2}}\)

\(\Leftrightarrow\dfrac{2-t}{t}=\sqrt{\dfrac{t^2-4}{2}}\Rightarrow\left(2-t\right)^2=t^2\left(\dfrac{t^2-4}{2}\right)\)

\(\Rightarrow2\left(2-t\right)^2=t^4-4t^2\Leftrightarrow-\left(t-2\right)\left(t^3+2t^2-2t+4\right)=0\Leftrightarrow t=2\left(tm\right)\Rightarrow\sqrt{x+1}+\sqrt{3-x}=2\Rightarrow4+2\sqrt{\left(x+1\right)\left(3-x\right)}=4\Leftrightarrow\left(x+1\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) thử lại \(\Rightarrow\)\(x=3;x=-1\) \(là\) \(nghiệm\)