Question

[Joy]

Giải pt:

\(\sqrt{x+5}+\sqrt{3-x}=2(\sqrt{15-2x-x^2}+1)\) 

Answer 1

[Nguyễn Phương Uyên]

\(đk:-5\le x\le3\)

\(\sqrt{x+5}+\sqrt{3-x}=2\left(\sqrt{15-2x-x^2}+1\right)\)                     (1)

\(\Leftrightarrow\sqrt{x+5}+\sqrt{3-x}=2\left(\sqrt{\left(x+5\right)\left(3-x\right)}+1\right)\)

\(\Leftrightarrow\sqrt{x+5}=\sqrt{3-x}=2\sqrt{\left(x+5\right)\left(3-x\right)}+2\)

đặt \(\sqrt{x+5}+\sqrt{3-x}=t\)   (đk t > 0)

\(\Leftrightarrow t^2=x+5+2\sqrt{\left(x+5\right)\left(3-x\right)}+3-x\)

\(\Leftrightarrow t^2=8+2\sqrt{\left(x+5\right)\left(3-x\right)}\)

\(\Leftrightarrow t^2=6+\left(2+2\sqrt{\left(x+5\right)\left(3-x\right)}\right)\)    và (1)

\(\Rightarrow t=t^2-6\)

\(\Leftrightarrow t^2-t-6=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=-2\left(loai\right)\\t=3\end{cases}}\)    

ta có : \(8+2\sqrt{\left(x+5\right)\left(3-x\right)}=3^2=9\)

\(\Leftrightarrow2\sqrt{\left(x+5\right)\left(3-x\right)}=1\)

\(\Leftrightarrow\sqrt{\left(x+5\right)\left(3-x\right)}=\frac{1}{2}\)

\(\Leftrightarrow\left(x+5\right)\left(3-x\right)=\frac{1}{4}\)

\(\Leftrightarrow59-8x-4x^2=0\)

\(\Leftrightarrow4x^2+8x+4-63=0\)

\(\Leftrightarrow4\left(x+1\right)^2=63\) \(\Leftrightarrow\left(x+1\right)^2=\frac{63}{4}\Leftrightarrow x+1=\pm\sqrt{\frac{63}{4}}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{63}{4}}-1\left(tm\right)\)