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a) A = B : C = \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]\). \(\frac{\sqrt{x^3y}+\sqrt{xy^3}}{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}\)
A xác định <=> x > 0 và y > 0
\(B=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]=\frac{2}{\sqrt{xy}}+\frac{1}{x}+\frac{1}{y}=\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\)
\(C=\frac{\sqrt{x}.\left(x+y\right)+\sqrt{y}.\left(x+y\right)}{\sqrt{xy}.\left(x+y\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right).\left(x+y\right)}{\sqrt{xy}.\left(x+y\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\)
=> A = B : C = \(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\) : \(\left(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\right)\) = \(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\)
c) \(A=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\ge2.\sqrt{\frac{1}{\sqrt{y}}.\frac{1}{\sqrt{x}}}=2.\sqrt{\frac{1}{\sqrt{6}}}\)
=> A nhỏ nhất bằng \(2.\sqrt{\frac{1}{\sqrt{6}}}\) khi \(\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{x}}\) => x = y = \(\sqrt{6}\)
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
\(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)
\(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1.\)
Vậy giá trị biểu thức \(A=1.\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\)
\(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\)
Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk )
=> \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\)
\(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\)
\(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\)
\(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\)
1. ĐKXĐ : \(xy>0\)
Ta có : \(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{-\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-2\sqrt{xy}+y+\sqrt{xy}}\right)\)
\(=\dfrac{\left(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(x+\sqrt{xy}+y\right)}{x-\sqrt{xy}+y}=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
2. Ta thấy : \(x-\sqrt{xy}+y=x-\dfrac{2.\sqrt{x}.\sqrt{y}}{2}+\dfrac{y}{4}+\dfrac{3y}{4}\)
\(=\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2+\dfrac{3y}{4}\)
Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-\dfrac{\sqrt{y}}{2}\right)^2\ge0\\\dfrac{3y}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow x-\sqrt{xy}+y\ge0\)
Lại có : \(\sqrt{xy}\ge0\)
\(\Rightarrow P\ge0\) ( ĐPCM )