Giải pt [tex]2x+\sqrt{3x^{2}+2x} = 3\sqrt{x}+3\sqrt{3x+2}+2\sqrt{2}-1[/tex]
ta có :
[tex](\sqrt{x}+\sqrt{3x+2})^{2}=x+3x+2+2\sqrt{x}.\sqrt{3x+2}=4x+2+2\sqrt{3x^{2}+2x}[/tex]
đặt
[tex]\sqrt{x}+\sqrt{3x+2}=t\Rightarrow t^{2}=4x+2+2\sqrt{3x^{2}+2x} \\\Rightarrow \frac{t^{2}-2}{2}=2x+\sqrt{3x^{2}+2x}=3\sqrt{x}+3\sqrt{3x+2}+2\sqrt{2}-1 \\=3(\sqrt{x}+\sqrt{3x+2})+2\sqrt{2}-1=3t+2\sqrt{2}-1 \\\Leftrightarrow t^{2}-6t-4\sqrt{2}=0[/tex]
đến đây dễ rồi !!!