Cho \(A=\sqrt{1-\dfrac{1}{xy}}\). Biết x, y \(\in\)Q và x\(\ne\)0; y \(\ne\)0 thỏa mãn : x3+ y3 = 2x2y2
Chứng minh rằng : A\(\in\)Q
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24 tháng 9 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
9 tháng 7 2022
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
AK
16 tháng 7 2018
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
7 tháng 10 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
10 tháng 7 2022
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
5 tháng 8 2018
\(A=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
31 tháng 7 2018
1) ta có : \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow P=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(\Leftrightarrow P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)
2) ta có : \(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(B=\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}-\dfrac{4}{2\sqrt{6}}+\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)^2-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+3\right)}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}.\dfrac{2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\sqrt{3}}=\dfrac{1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{6}-\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)