ĐKXĐ: \(x\ge0\)
\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(x+1\right)^2\left(2x+1\right)}=\left(x+1\right)\left(5x^2-8x+8\right)\)
\(\Leftrightarrow2\left(x+1\right)\sqrt{x}+\left(x+1\right)\sqrt{3\left(2x+1\right)}=\left(x+1\right)\left(5x^2-8x+8\right)\)
Do \(x\ge0\Rightarrow x+1>0\) chia cả 2 vế cho \(x+1\) ta được:
\(2\sqrt{x}+\sqrt{3\left(2x+1\right)}=5x^2-8x+8\)
\(\Leftrightarrow5x^2-10x+5+x+1-2\sqrt{x}+x+2-\sqrt{6x+3}=0\)
\(\Leftrightarrow5\left(x-1\right)^2+\left(\sqrt{x}-1\right)^2+\frac{\left(x-1\right)^2}{x+2+\sqrt{6x+3}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\left(5\left(\sqrt{x}+1\right)^2+1+\frac{\left(\sqrt{x}+1\right)^2}{x+2+\sqrt{6x+3}}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\) (ngoặc phía sau luôn dương)
\(\Leftrightarrow x=1\)
sửa đề tí nha: dấu "=" đầu tiên thay thành dấu "+"
