Ta có: \(\frac{x^3-2x^2+3x-6}{5x-10}=\frac{x^2\left(x-2\right)+3\left(x-2\right)}{5\left(x-2\right)}=\frac{\left(x^2+3\right)\left(x-2\right)}{5\left(x-2\right)}=\frac{x^2+3}{5}\)
Vì \(x^2\ge0\forall x\Rightarrow x^2+3\ge3>0\forall x\)
\(\Rightarrow\frac{x^2+3}{5}>0\forall x\Rightarrowđpcm\)
x2(x-2)+3(x-2) / 5(x-2)
x2 +3 / 5
mà ta có tử luôn luôn khác 0
x2≥0⇒ x2+3≥3 >2