\(\Leftrightarrow\sqrt{\dfrac{1}{2}-x^2}\cdot4x\cdot\left(1-x\right)=1-x\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}-x^2}=\dfrac{1}{4x}\)
\(\Leftrightarrow\dfrac{1}{2}-x^2=\dfrac{1}{16x^2}\)
\(\Leftrightarrow\dfrac{8x^2}{16x^2}-\dfrac{16x^4}{16x^2}=\dfrac{1}{16x^2}\)
=>-16x^4+8x^2=1
=>16x^4-8x^2+1=0
=>(4x^2-1)=0
=>x=1/2 hoặc x=-1/2