Question

Giải phương trình: \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)

Answer 1

\(ĐK:x\ge\dfrac{-1}{4}.\)Khi đó phương trình tương đương với:
\(x+\sqrt{x+\dfrac{1}{4}+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{4}}=2\)\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)\(\Leftrightarrow x+\left|\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right|=2\)\(\Leftrightarrow x+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=2\)

\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)\(\Leftrightarrow\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=\sqrt{2}\)\(\Leftrightarrow\sqrt{x+\dfrac{1}{4}}=\sqrt{2}-\dfrac{1}{2}\)\(\Leftrightarrow x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)\(\Leftrightarrow x=2-\sqrt{2}\)
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