Kết quả tìm kiếm

$P = x + \dfrac{1}{2}y - \sqrt{xy} - 3\sqrt{x} + \sqrt{y} + 4\\ 2P = 2x + y - 2\sqrt{xy} - 6\sqrt{x} + 2\sqrt{y} + 8\\ 2P = (x - 2\sqrt{xy} + y) - (2\sqrt{x} - 2\sqrt{y}) + (x - 4\sqrt{x} + 4) + 4\\ 2P = (\sqrt{x} - \sqrt{y})^2 - 2(\sqrt{x} - \sqrt{y}) + 1 + (\sqrt{x} - 2)^2 + 3\\ 2P = (\sqrt{x}...

Hướng dẫn a) Xét $\Delta AMN \sim \Delta ACB \Rightarrow AM.AB = AN.AC$ b)Ta có : $MA . MB+NA . NC = MH^2 + NH^2 = MN^2 = AH^2 = BH . CH$ c) $S_{ABH} = \dfrac{AH.BH}{2}\\ S_{ACH} = \dfrac{AH.CH}{2}\\ \Rightarrow \dfrac{S_{ABH}}{S_{ACH}} = \dfrac{BH}{CH}$ mà $\Delta ABH\sim \Delta ACH$ nên...

Do $\Delta ABC\sim \Delta DEF$ nên $\dfrac{AB}{DE} = \dfrac{BC}{EF} \Leftrightarrow \dfrac{AB}{BC} = \dfrac{DE}{EF} \Leftrightarrow \dfrac{4}{5} = \dfrac{12}{EF} \Leftrightarrow EF = 15$ $\dfrac{AB}{DE} = \dfrac{AC}{DF} \Leftrightarrow \dfrac{DF}{DE} = \dfrac{AC}{AB} = \dfrac{AC - DF}{AB-DE}...

a) $m - \sqrt{m} = \sqrt{m}(\sqrt{m} - 1)$ Do $m > 1$ nên $\sqrt{m} - 1 > 0$ và $\sqrt{m} >0$ Suy ra : $m - \sqrt{m} > 0$ Hay $m > \sqrt{m}$ b) Tương tự câu a)