Question

Giải pt:

\(\sqrt{2x-2}-\sqrt{6x-9}=16x^2-48x+35\)

Answer 1

ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(16x^2-48x+35+\left(\sqrt{6x-9}-\sqrt{2x-2}\right)=0\)

\(\Leftrightarrow\left(4x-7\right)\left(4x-5\right)+\dfrac{4x-7}{\sqrt{6x-9}+\sqrt{2x-2}}=0\)

\(\Leftrightarrow\left(4x-7\right)\left(4x-5+\dfrac{1}{\sqrt{6x-9}+\sqrt{2x-2}}\right)=0\)

\(\Leftrightarrow4x-7=0\)