Question

Giải Bất Phương Trình

\(\frac{-6}{x^2-4x+5}\ge x^2-4x+1\)

Help gấp :DDD

Answer 1

\(\Rightarrow-6>=\left(x^2-4x+1\right)\left(x^2-4x+5\right)=\left(x^2-4x\right)^2+5\left(x^2-4x\right)+x^2-4x+5\)

\(=\left(x^2-4x\right)^2+6\left(x^2-4x\right)+9-4=\left(x^2-4x\right)^2+2\cdot3\left(x^2-4x\right)+3^2-4\)

\(=\left(x^2-4x+3\right)^2-4\)

\(\Rightarrow\left(x^2-4x+3\right)^2-4< =-6\Rightarrow\left(x^2-4x+3\right)^2< =-2\)

vì \(\left(x^2-4x+3\right)^2>=0\Rightarrow\left(x^2-4x+3\right)^2< =-2\)vô lí \(\Rightarrow x\in\varnothing\)