Tìm GTNN của $\dfrac{2x+ 4}{3\sqrt{x – 2}}$ khi x $\geq$ 0 24/11/2021 Bởi Savannah Tìm GTNN của $\dfrac{2x+ 4}{3\sqrt{x – 2}}$ khi x $\geq$ 0
`ĐKXĐ: x>2` `A=(2x+4)/(3\sqrt(x-2))` `A=[8/3(3\sqrt(x-2))+2(x-2-4\sqrt(x-2)+4)]/(3\sqrt(x-2))` `A=8/3+2(\sqrt(x-2)-2)^2/(3\sqrt(x-2))≥8/3` Dấu `=` xảy ra `⇔2(\sqrt(x-2)-2)^2/(3\sqrt(x-2))=0⇔x=6` Vậy $Min_A=\dfrac{8}{2}⇔x=6$ Bình luận
`ĐKXĐ: x>2`
`A=(2x+4)/(3\sqrt(x-2))`
`A=[8/3(3\sqrt(x-2))+2(x-2-4\sqrt(x-2)+4)]/(3\sqrt(x-2))`
`A=8/3+2(\sqrt(x-2)-2)^2/(3\sqrt(x-2))≥8/3`
Dấu `=` xảy ra `⇔2(\sqrt(x-2)-2)^2/(3\sqrt(x-2))=0⇔x=6`
Vậy $Min_A=\dfrac{8}{2}⇔x=6$