Question

Giải PT: \(\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}\)

Answer 1

\(\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}\)

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

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{x-1}{x}=0\)

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

Pt \(\dfrac{1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{1}{x}=0\) vô n₀

=> x - 1 = 0

<=> x = 1 (nhận)