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a) Ta có: \(C=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)

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

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

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

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

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

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

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\y>0\\x\ne y\end{matrix}\right.\)

Ta có: \(C-1=\frac{x-\sqrt{xy}+y}{\sqrt{xy}}-1\)

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

\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\forall x,y\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow C-1>0\)

hay C>1(đpcm)

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 :\(\hept{\begin{cases}x>0\\y>0\\x\ne y\end{cases}}\)

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

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

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

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

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

_Minh ngụy_

11 tháng 8 2017

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

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

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

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

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

\(=\sqrt{y}-\sqrt{x}\)

29 tháng 7 2017

Thưa....bạn.....mình....chịu.....

16 tháng 8 2017

Ê bạn... thiên vị ak.

Sao ko đợi người nào giỏi trả lời