Rút gọn: \(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y}\right).\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{x\sqrt{x}+y\sqrt{y}}\)
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BT
7 tháng 8 2017
\(A=\left\{\frac{2\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{x}\left(x+y\right)}{\sqrt{x}}\right\}.\left(\frac{\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right)^2.\)
=> \(A=\left(2\sqrt{xy}+x+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
=> \(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}=1\)
ĐS: A=1
15 tháng 8 2017
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
15 tháng 8 2017
a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
24 tháng 7 2017
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
10 tháng 8 2017
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
9 tháng 5 2018
ĐkXĐ \(x\ge0,y\ge0\)
Ta có \(A=\left(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right).\frac{1}{\left(x-y\right)}+\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(x-2\sqrt{xy}+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right)+\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\left(3\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\)
vậy với...... thì biểu thức đã cho đc rút gọn là ...
6 tháng 9 2020
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)
a) \(C=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)
\(C=\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(C=\frac{x+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}\)
\(C=\frac{\left(x+y-\sqrt{xy}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x\sqrt{y}-y\sqrt{x}}\)
\(C=\frac{\left(x+y-\sqrt{xy}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}\)
\(C=\frac{x+y-\sqrt{xy}}{\sqrt{xy}}\)
b)Giả sử \(C>1\)
\(\Leftrightarrow\frac{x+y-\sqrt{xy}}{\sqrt{xy}}>1\)
\(\Leftrightarrow\frac{x+y-\sqrt{xy}-\sqrt{xy}}{\sqrt{xy}}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\)( luôn đúng với mọi \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\))
6 tháng 9 2020
Nhầm ĐKXĐ :\(\hept{\begin{cases}x>0\\y>0\\x\ne y\end{cases}}\)
1 tháng 2 2022
\(=\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x^2-y^2\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}\)
\(=\dfrac{\sqrt{xy}\left(x+y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}=-\sqrt{x}+\sqrt{y}\)(1)
Khi x=3 và \(y=4+2\sqrt{3}\) vào (1), ta được:
\(=-\sqrt{3}+\sqrt{4+2\sqrt{3}}=-\sqrt{3}+\sqrt{3}+1=1\)
Lời giải:
ĐK: \(x,y\geq 0; x\neq y\). Để cho gọn đặt \(\sqrt{x}=a; \sqrt{y}=b\). Khi đó:
\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y}\right).\frac{(\sqrt{x}-\sqrt{y})^2}{x\sqrt{x}+y\sqrt{y}}\)
\(=(\frac{a^2-b^2}{a-b}-\frac{a^3-b^3}{a^2-b^2}).\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{(a^2-b^2)(a+b)-(a^3-b^3)}{a^2-b^2}.\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{ab(a-b)}{(a-b)(a+b)}.\frac{(a-b)^2}{a^3+b^3}=\frac{ab(a-b)^2}{(a+b)(a^3+b^3)}\)
\(=\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})(x\sqrt{x}+y\sqrt{y})}\)