Question

Giải phương trình ;

a) \(\sqrt{x}+\sqrt{x+1}=\dfrac{1}{\sqrt{x}}\)

b) \(x+\dfrac{2\sqrt{2}x}{\sqrt{1}+x}=1\)

Answer 1

a) \(\sqrt{x}+\sqrt{x+1}=\dfrac{1}{\sqrt{x}}\)

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

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

\(\Leftrightarrow x+\sqrt{x^2+x}=1\)

\(\Leftrightarrow\sqrt{x^2+x}=1-x\)

ĐKXĐ: \(1-x\ge0\Rightarrow x\ge1\)

\(\Leftrightarrow x^2+x=\left(1-x\right)^2\)

\(\Leftrightarrow x^2+x=1-2x+x^2\)

\(\Leftrightarrow x^2-x^2+x+2x=1\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\) (nhận)

\(S=\dfrac{1}{3}\)