ĐK: \(-3\le x\le2\)
\(4\left(x+1\right)\left(\sqrt{x+3}-\sqrt{2-x}\right)=-x^2+12x+13\)
<=> \(4\left(x+1\right)\left(\sqrt{x+3}-\sqrt{2-x}\right)+\left(x+1\right)\left(x-13\right)=0\)
<=> \(\left(x+1\right)\left[4\left(\sqrt{x+3}-\sqrt{2-x}\right)+x-13\right]=0\)
<=> \(\orbr{\begin{cases}x+1=0\left(1\right)\\4\left(\sqrt{x+3}-\sqrt{2-x}\right)+x-13=0\left(2\right)\end{cases}}\)
(1) <=> x = - 1 ( thỏa mãn )
(2) <=> \(4\left(\sqrt{x+3}-\sqrt{2-x}\right)=13-x\)
Ta có VT \(\le4\sqrt{x+3+2-x}=4\sqrt{5}\)với \(-3\le x\le2\)
\(VP\ge11\)với \(-3\le x\le2\)
=> VP > VT mọi \(-3\le x\le2\)
pt (2) vô nghiệm
Vậy x = - 1 là nghiệm.
