Question

\(\sqrt{x^2-1}+\sqrt{x^2-3x+2}\le2\sqrt{x^2-x}\)

Answer 1

đkxđ:\(\left[ \begin{array}{l}x \geq 2\\x \leq -1\end{array} \right.\) 

`bpt<=>\sqrt{x-1}(\sqrt{x+1}+\sqrt{x-1}-2\sqrtx)<=0`

Vì `\sqrt{x-1}>=1>0`

`=>\sqrt{x+1}+\sqrt{x-1}-2\sqrtx<=0`

`<=>\sqrt{x+1}+\sqrt{x-1}<=2\sqrtx`

BP 2 vế

`=>2x+2\sqrt{x^2-1}<=4x`

`<=>>\sqrt{x^2-1}<=x`

`<=>x^2-1<=x^2`(luôn đúng)

Vậy với \(\left[ \begin{array}{l}x \geq 2\\x \leq -1\end{array} \right.\)  thì.......