ĐKXĐ: x>0
Ta có: \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
=>\(\sqrt{x^2+3x}+\sqrt{4x+8}=2x+\sqrt{\frac{x^2+5x+6}{x}}\)
=>\(2x-4+\sqrt{\frac{x^2+5x+6}{x}}-\sqrt{10}=\sqrt{x^2+3x}-\sqrt{10}+\sqrt{4x+8}-4\)
=>\(2\left(x-2\right)+\frac{\left(\frac{x^2+5x+6}{x}-10\right)}{\sqrt{\frac{x^2+5x+6}{x}}+\sqrt{10}}=\frac{x^2+3x-10}{\sqrt{x^2+3x}+10}+\frac{4x+8-16}{\sqrt{4x+8}+4}\)
=>\(2\left(x-2\right)+\frac{\left(\frac{x^2+5x+6}{x}-\frac{10x}{x}\right)}{\sqrt{\frac{x^2+5x+6}{x}}+\sqrt{10}}=\frac{x^2+5x-2x-10}{\sqrt{x^2+3x}+10}+\frac{4x-8}{\sqrt{4x+8}+4}\)
=>\(2\left(x-2\right)+\frac{\frac{x^2-5x+6}{x}}{\sqrt{\frac{x^2+5x+6}{x}}+\sqrt{10}}=\frac{\left(x+5\right)\left(x-2\right)}{\sqrt{x^2+3x}+10}+\frac{4\left(x-2\right)}{\sqrt{4x+8}+4}\)
=>\(2\left(x-2\right)+\frac{\left(x-2\right)\left(x-3\right)}{x\left(\sqrt{\frac{x^2+5x+6}{x}}+\sqrt{10}\right)}-\frac{\left(x+5\right)\left(x-2\right)}{\sqrt{x^2+3x}+10}-\frac{4\left(x-2\right)}{\sqrt{4x+8}+4}=0\)
=>\(\left(x-2\right)\cdot\left\lbrack2+\frac{x-3}{x\cdot\left(\sqrt{\frac{x^2+5x+6}{x}}+\sqrt{10}\right)}-\frac{x+5}{\sqrt{x^2+3x}+10}-\frac{4}{\sqrt{4x+8}+4}\right\rbrack=0\)
=>x-2=0
=>x=2(nhận)
