\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-1\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}+1\right)\)
Rút gọn D
K
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22 tháng 10 2016
a/ Bạn tự tìm ĐKXĐ
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
Xét
- \(=\frac{\left(\sqrt{x}+1\right)\left(1-\sqrt{xy}\right)+\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{\sqrt{x}-x\sqrt{y}+1-\sqrt{xy}+xy+\sqrt{xy}+x\sqrt{y}+\sqrt{x}+1-xy}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{2\sqrt{x}+2}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
- \(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)
\(=\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{-2\sqrt{xy}-2x\sqrt{y}}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}=\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(\Rightarrow A=\frac{2\left(\sqrt{x}+1\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}:\frac{2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}=\frac{1}{\sqrt{xy}}\)
b/ Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) với \(a=\frac{1}{\sqrt{x}},b=\frac{1}{\sqrt{y}}\) được :
\(A=\frac{1}{\sqrt{x}.\sqrt{y}}\le\frac{1}{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2=\frac{1}{4}.6^2=9\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\end{cases}}\Leftrightarrow x=y=\frac{1}{9}\)
Vậy ........................................................
10 tháng 1 2019
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
11 tháng 8 2019
\(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
+) Đặt \(Q=\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)
\(Q=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}+\frac{xy-1}{xy-1}\)
\(Q=\frac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}+xy-1}{xy-1}\)
\(Q=\frac{-2-2\sqrt{x}}{xy-1}\)
\(Q=\frac{-2\left(\sqrt{x}+1\right)}{xy-1}\)
+) Đặt \(K=1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)
\(K=\frac{xy-1}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}\)
\(K=\frac{xy-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{xy-1}\)
\(K=\frac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)
\(K=\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)
Ta có : \(P=Q:K\)
\(\Leftrightarrow P=\frac{-2\left(\sqrt{x}+1\right)}{xy-1}:\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)
\(\Leftrightarrow P=\frac{-2\left(\sqrt{x}+1\right)\left(xy-1\right)}{-2\sqrt{xy}\left(\sqrt{x}+1\right)\left(xy-1\right)}\)
\(\Leftrightarrow P=\frac{1}{\sqrt{xy}}\)
Vậy...
10 tháng 8 2017
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\)
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\)
\(D=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}-1\right)^2}\)
\(D=\left(\sqrt{x}+1\right)^2\left(\frac{1}{\left(\sqrt{xy}+1\right)^2}-\frac{1}{\left(\sqrt{xy}-1\right)^2}\right)\)
\(D=\left(\sqrt{x}+1\right)^2\cdot\frac{xy+1-2\sqrt{xy}-xy-1-2\sqrt{xy}}{\left(xy-1\right)^2}\)
\(D=\frac{\left(\sqrt{x}+1\right)^2\cdot\left(-4\sqrt{xy}\right)}{\left(xy-1\right)^2}\)