\(x^2-x\left(4x^3-x+3\right)+\left(4x^3-x+3\right)^2+\frac{1}{2}=0.\)
\(\left\{x-\frac{\left(4x^3-x+3\right)}{2}\right\}^2+\frac{4\left(4x^3-x+3\right)^2-\left(4x^3-x+3\right)+2}{4}=0\)
đặt \(4x^3-x+3=t\)
\(\left(x-\frac{t}{2}\right)^2+\frac{4t^2-t+2}{4}=0\)
\(4t^2-t+2=pain\)
\(\Delta pain=1^2-4.2.4< 0\)
vậy suy ra vậy thằng Pain luôn dương youbtobe đã nói vậy
\(\left(x-\frac{t}{2}\right)^2\ge0\)
\(\frac{4t-t+2}{4}>0\)
suy ra vô nghiệm muahahahahaha
\(\left(x-\frac{t}{2}\right)^2=-\frac{\left(4t^2-t-2\right)}{4}\)
mik ko hỉu bạn