bài này dùng liên hợp nhé bạn
pt<=> \(x^2+3x-\sqrt{2+2x}-\sqrt{5-x}=0\)
<=> \(x+1-\sqrt{2+2x}+x+1-\sqrt{5-x}+x^2+x-2=0\)
<=> \(\frac{x^2+2x+1-2-2x}{x+1+\sqrt{2+2x}}+\frac{x^2+2x+1-5+x}{x+1+\sqrt{5-x}}+\left(x-1\right)\left(x+2\right)=0\)
<=> \(\frac{x^2-1}{x+1+\sqrt{2+2x}}+\frac{x^2+3x-4}{x+1+\sqrt{5-x}}+\left(x-1\right)\left(x+2\right)=0\)
<=> \(\frac{\left(x-1\right)\left(x+1\right)}{x+1+\sqrt{2+2x}}+\frac{\left(x-1\right)\left(x+4\right)}{x+1+\sqrt{5-x}}+\left(x-1\right)\left(x+2\right)=0\)
<=>: \(\left(x-1\right)\left[\frac{x+1}{x+1+\sqrt{2+2x}}+\frac{x+4}{x+1+\sqrt{5-x}}+\left(x+2\right)\right]=0\)
==> x=1
còn cái trong căn thì khác 0
vậy x=1
