Question

tìm x:4-x=2(x-4)^2

 

Answer 1

\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow-2\left(x-4\right)^2=x-4\)

\(\Rightarrow-2\left(x-4\right)=1\)

\(\Rightarrow x=\dfrac{7}{2}\)

Answer 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

Answer 3

\(x:4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow x:4-x=2\left(x^2+16-8x\right)\)

\(\Leftrightarrow x:4-x=2x^2+32-16x\)

\(\Leftrightarrow x:4=2x^2+32-15x\)

\(\Leftrightarrow x=8x^2+128-60x\)

\(\Leftrightarrow0=8x^2+128-61x\)(VN)

Vậy không có x thỏa mãn