Question

\(2x+\sqrt{x+\sqrt{x-\dfrac{1}{4}}}=2\) giải pt giúp mik nha

Answer 1

ĐKXĐ: \(1\ge x\ge\dfrac{1}{4}\)

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

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

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

\(\Leftrightarrow\sqrt{x-\dfrac{1}{4}}=\dfrac{3}{2}-2x\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{9}{4}-6x+4x^2\)

\(\Leftrightarrow4x^2-7x+\dfrac{5}{2}=0\)

\(\Leftrightarrow4\left(x-\dfrac{5}{4}\right)\left(x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\left(L\right)\\x=\dfrac{1}{2}\left(TM\right)\end{matrix}\right.\)