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13 tháng 7 2017

\(P=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}.\)

\(P=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1+x+y+xy}\)

\(P=\frac{2\sqrt{x}}{1+x+y+xy}\)Với ĐK \(x\ge0\) và \(y\ge0\)Và \(xy\ne1\)

Nguyễn Ngọc Anh Minh bạn làm sai rồi kìa bước cuối cùng vẫn còn \(2y\sqrt{x}\)

14 tháng 7 2017

a, ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(Q=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

b. \(Q>1\Rightarrow Q-1>0\Rightarrow\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Rightarrow\frac{x+2}{\sqrt{x}-1}>0\)

TH1 \(\hept{\begin{cases}x+2>0\\\sqrt{x}-1>0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x>1\end{cases}\Rightarrow}x>1}\)

TH2 \(\hept{\begin{cases}x+2< 0\\\sqrt{x}-1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\0\le x< 1\end{cases}\left(l\right)}}\)

Vậy \(x>1\)thì \(Q>1\)

a) Ta có: \(P=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+2xy+y}{1-xy}\right)\)

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

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

\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)

5 tháng 7 2021

Đk:\(xy\ne1;x\ge0;y\ge0\)

 \(P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1-xy+x+y+2xy}{1-xy}\)

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

\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{\left(1+x\right)\left(1+y\right)}{1-xy}\)\(=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+x\right)\left(1+y\right)}=\dfrac{2\sqrt{x}}{1+x}\)

b) Áp dụng AM-GM có:

\(1+x\ge2\sqrt{x}\Leftrightarrow\)\(\dfrac{2\sqrt{x}}{1+x}\le1\)

Dấu "=" xảy ra khi x=1 (tm)

Vậy \(P_{max}=1\)

5 tháng 7 2017

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

\(=\left[\left(\sqrt{x}+\sqrt{y}\right)-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right].\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

5 tháng 7 2017

Mình gi rút gọn bạn tự hiểu nha:

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

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

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

=

12 tháng 10 2022

a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)

b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)

\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)

\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)

c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

27 tháng 10 2019

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 \(x=y=4\)