BĐT cosi : Tìm GTLN: A=(2-x)(1+2x) B= $\frac{\sqrt[]{y-2} }{y}$ + $\frac{\sqrt[]{x-3} }{x}$

Giải thích các bước giải:

a.$A=(2-x)(1+2x)=\dfrac{1}{2}.(4-2x)(1+2x)\le \dfrac{1}{2}.(\dfrac{4-2x+1+2x}{2})^2=\dfrac{25}{8}$

Dấu = xảy ra khi $4-2x=1+2x\rightarrow x=\dfrac{3}{4}$ 

b.Ta có :

$\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{(y-2).2}}{y\sqrt{2}}\le \dfrac{\sqrt{\dfrac{(y-2+2)^2}{4}}}{y\sqrt{2}}=\dfrac{y}{y.2\sqrt{2}}=\dfrac{1}{2\sqrt{2}}$

$\dfrac{\sqrt{x-3}}{x}=\dfrac{\sqrt{(x-3).3}}{x\sqrt{3}}\le \dfrac{\sqrt{\dfrac{(x-3+3)^2}{4}}}{x\sqrt{3}}=\dfrac{x}{x.2\sqrt{3}}=\dfrac{1}{2\sqrt{3}}$

$\rightarrow \dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{x-3}}{x}\le \dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}$

Dấu = xảy ra khi $x=4,y=6$