\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow-2\left(x-4\right)^2=x-4\)
\(\Rightarrow-2\left(x-4\right)=1\)
\(\Rightarrow x=\dfrac{7}{2}\)
Ta có: \(\dfrac{1}{4}x-x=2\left(x-4\right)^2\)
\(\Leftrightarrow2x^2-16x+32+\dfrac{3}{4}x=0\)
\(\Leftrightarrow2x^2-\dfrac{61}{4}x+32=0\)
\(\text{Δ}=\left(-\dfrac{61}{4}\right)^2-4\cdot2\cdot32=-\dfrac{375}{16}< 0\)
Vì Δ<0 nên phương trình vô nghiệm
\(x:4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow x:4-x=2\left(x^2+16-8x\right)\)
\(\Leftrightarrow x:4-x=2x^2+32-16x\)
\(\Leftrightarrow x:4=2x^2+32-15x\)
\(\Leftrightarrow x=8x^2+128-60x\)
\(\Leftrightarrow0=8x^2+128-61x\)(VN)
Vậy không có x thỏa mãn
