- ĐK : \(x\ge\dfrac{-1}{4}\)
- Ta có : \(x+\sqrt{x+\dfrac{1}{2}\sqrt{x+\dfrac{1}{4}}}=2\) (*)
Đặt \(\sqrt{x+\dfrac{1}{4}}=a\left(a\ge0\right)\) => \(x=a^2-\dfrac{1}{4}\)
Khi đó : (*) <=> \(\left(x+\dfrac{1}{4}\right)-\dfrac{9}{4}+\sqrt{a^2+\dfrac{1}{2}a-\dfrac{1}{4}}=0\)
<=> \(a^2-\dfrac{9}{4}+\sqrt{a^2+\dfrac{1}{2}-\dfrac{1}{4}}\)= 0
<=> \(\left(a^2+\dfrac{1}{2}a-\dfrac{1}{4}\right)+2\sqrt{a^2+\dfrac{1}{2}a-\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}=0\)
<=> \(\left(\sqrt{a^2+\dfrac{1}{2}a-\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\)
Vì \(\left(\sqrt{a^2+\dfrac{1}{2}a-\dfrac{1}{4}}+\dfrac{1}{2}\right)>0\)
=> \(\sqrt{a^2+\dfrac{1}{2}a-\dfrac{1}{4}}=\dfrac{1}{6}\)
=> \(a^2+\dfrac{1}{2}a=\dfrac{1}{36}+\dfrac{1}{4}=\dfrac{5}{18}\)
=> \(a^2+\dfrac{1}{2}a-\dfrac{5}{18}=0\)
<=> \(\left(a-\dfrac{1}{3}\right)\left(a+\dfrac{5}{6}\right)=0\)
=> \(\left[{}\begin{matrix}a=-\dfrac{5}{6}< 0\left(loại\right)\\a=\dfrac{1}{3}\left(t.m\right)\end{matrix}\right.\)
Với a=\(\dfrac{1}{3}\) => \(x+\dfrac{1}{4}\) = \(\dfrac{1}{9}\) => x= \(\dfrac{-5}{36}\) > \(\dfrac{-1}{4}\) (t.m)
Vậy -.
