Question

giải phương trình vô tỷ

\(\sqrt{2x-3}\) + 6 = 2x +\(\sqrt{x}\)

Answer 1

ĐKXĐ: \(x\ge\dfrac{3}{2}\).

\(\sqrt{2x-3}+6=2x+\sqrt{x}\Leftrightarrow\sqrt{2x-3}-\sqrt{x}=2\left(x-3\right)\Leftrightarrow\dfrac{x-3}{\sqrt{2x-3}+\sqrt{x}}-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tmđkxđ\right)\\\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}=2\left(1\right)\end{matrix}\right.\)

Vì \(x\ge\dfrac{3}{2}\Rightarrow\sqrt{2x-3}+\sqrt{x}\ge\sqrt{\dfrac{3}{2}}>\dfrac{1}{2}\)nên (1) vô lí

Vậy pt đã cho có nghiệm là x=3