)
bài này có cái căn uy hiếp chứ chả ích j
\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow\sqrt{x+\frac{1}{4}+2\sqrt{x+\frac{1}{4}}\cdot\frac{1}{2}+\frac{1}{4}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right|=2\)
\(\Leftrightarrow\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\) (do \(\sqrt{x+\frac{1}{4}}+\frac{1}{2}>0\forall x\))
\(\Leftrightarrow\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{9}{4}\)
\(\Leftrightarrow x=2\)
\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
Đặt \(t=\sqrt{x+\frac{1}{4}}\Rightarrow t^2=x+\frac{1}{4}\Rightarrow x=t^2-\frac{1}{4}\)
\(\Leftrightarrow t^2-\frac{1}{4}+\sqrt{t^2+t+\frac{1}{4}}=2\)
\(\Leftrightarrow\sqrt{t^2+t+\frac{1}{4}}=\frac{9}{4}-t^2\)
\(\Leftrightarrow t^2+t+\frac{1}{4}=\left(\frac{9}{4}-t^2\right)^2\)
\(\Leftrightarrow t^4-\frac{11}{2}t^2-t+\frac{77}{16}=0\)
\(\Leftrightarrow16t^4-88t^2-16t+77=0\)
\(\Leftrightarrow\left(4t^2-4t-11\right)\left(4t^2+4t-7\right)=0\)
