Question

[nguyễn hương giang]

\(4\left(x^2+1\right)=3\sqrt{2x^2-7x+3}+14x\)

Answer 1

[Vũ Tiền Châu]

Ta có pt \(4x^2-14x-2=3\left(\sqrt{2x^2-7x+3}-2\right)\Leftrightarrow2\left(2x^2-7x-1\right)=3.\dfrac{2x^2-7x+3-4}{\sqrt{2x^2-7x+3}+2}\Leftrightarrow2\left(2x^2-7x-1\right)=3.\dfrac{2x^2-7x-1}{\sqrt{2x^2-7x+3}+2}\)

\(\Leftrightarrow\left(2x^2-7x-1\right)\left(2-\dfrac{3}{\sqrt{2x^2-7x+3}+2}\right)=0\)

Mà \(\sqrt{2x^2-7x+3}+2\ge2\Rightarrow\dfrac{3}{\sqrt{2x^2-7x+3}+2}\le\dfrac{3}{2}\Rightarrow2-\dfrac{3}{\sqrt{2x^2-7x+3}+2}>0\)

=> pt <=> \(2x^2-7x-1=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7-\sqrt{57}}{4}\\x=\dfrac{7+\sqrt{57}}{4}\end{matrix}\right.\)