chịu thua vô điều kiện xin lỗi nha : v

muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v

\(A=\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]:\frac{\sqrt{x^3}+y.\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(\Leftrightarrow A=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{x+y}{xy}\right]:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{2\sqrt{xy}+x+y}{xy}:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{xy}\left(x+y\right)}{xy\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Leftrightarrow A=\frac{\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

sai sót chỗ nào chỉ cho mk nhé. ý kia chốc nx làm nốt

a) \(P=\dfrac{\left(x^2+2xy+9y^2\right)-\left(x+3y-2\sqrt{xy}\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x^2+6xy+9y^2\right)-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x+3y\right)^2-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x+3y\right)\left(x+3y-2\sqrt{xy}\right)}{x+3y-2\sqrt{xy}}\)

\(P=x+3y\)

b) \(\dfrac{P}{\sqrt{xy}+y}=\dfrac{x+3y}{\sqrt{xy}+y}=\dfrac{\left(x+3y\right):y}{\left(\sqrt{xy}+y\right):y}=\dfrac{\dfrac{x}{y}+3}{\sqrt{\dfrac{x}{y}}+1}\)

Đặt \(t=\sqrt{\dfrac{x}{y}}>0\) và \(\dfrac{P}{\sqrt{xy}+y}=Q\) thì \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)

\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)

ĐK: \(x;y\ge0\) đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(P=a^2-2ab+3b^2-2a+1=3\left(\frac{1}{9}a^2-\frac{2}{3}ab+b^2\right)+\frac{2}{3}\left(a^2-3a+\frac{9}{4}\right)-\frac{1}{2}\)

\(P=3\left(\frac{a}{3}-b\right)^2+\frac{2}{3}\left(a-\frac{3}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)

\(\Rightarrow P_{min}=-\frac{1}{2}\) khi \(\left\{{}\begin{matrix}a=\frac{3}{2}\\b=\frac{1}{2}\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x=\frac{9}{4}\\y=\frac{1}{4}\end{matrix}\right.\)

a.\(DK:x,y>0\)

Ta co:

\(A=\frac{x+y+2\sqrt{xy}}{xy}.\frac{\sqrt{xy}\left(x+y\right)}{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

b.

Ta lai co:

\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\ge\frac{2\sqrt{\sqrt{x}.\sqrt{y}}}{4}=1\)

Dau '=' xay ra khi \(x=y=4\)

Vay \(A_{min}=1\)khi \(x=y=4\)

\(A=x-2\sqrt{xy}+3y-2\sqrt{x}+1=\left(x+y+1-2\sqrt{xy}-2\sqrt{x}+2\sqrt{y}\right)+\left(2y-2\sqrt{y}\right)\)

\(=\left(-\sqrt{x}+\sqrt{y}+1\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)

\(\Rightarrow MinA=-\frac{1}{2}\Leftrightarrow\hept{\begin{cases}\sqrt{y}-\sqrt{x}+1=0\\\sqrt{y}-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{1}{4}\end{cases}}\)