Toán 9 Giải phương trình

[tex](4x-1)^2(x^2+1)=(2x^2+2x+1)^2<=>x(4x-3)(4x^2+3)=0[/tex]

$\left (4x-1\right )\left (\sqrt{x^{2}+1}-\frac{5}{3}\right )=2x^{2}-\frac{14}{3}x+\frac{8}{3}\\\\\Leftrightarrow \left (4x-1\right )\left (\sqrt{x^{2}+1}-\frac{5}{3}\right )=2\left (x-\frac{4}{3}\right )\left (x-1\right )\\\\\Leftrightarrow \left (4x-1\right )\left (\frac{x^{2}+1-\frac{25}{9}}{\sqrt{x^{2}+1}+\frac{5}{3}}\right )=2\left (x-\frac{4}{3}\right )\left (x-1\right )\\\\\Leftrightarrow \left (4x-1\right )\left (x-\frac{4}{3}\right )\left (x+\frac{4}{3}\right ).\frac{1}{\sqrt{x^{2}+1}+\frac{5}{3}}=2\left (x-\frac{4}{3}\right )\left (x-1\right )$