cho \(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
tính giá trị bt: \(A=\dfrac{xy-1}{x+y}-\dfrac{1-xy}{2x-y}\)
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21 tháng 5 2022
a: \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{4xy}{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
b: \(=\sqrt{x}+\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)-2\sqrt{y}\)
\(=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0\)
c: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
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: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
28 tháng 5 2022
a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)
\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)
\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)
b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)
c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)
ND
24 tháng 7 2018
\(a.R=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}\right)\)
\(R=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+xy-1}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]:\left[\dfrac{xy-1-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)
\(R=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}+xy-1}{xy-1}:\dfrac{xy-1-x\sqrt{y}+\sqrt{x}+\sqrt{xy}+1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}}{xy-1}\)
\(R=\dfrac{-2\sqrt{x}-2}{xy-1}:\dfrac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)
\(R=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}.\dfrac{xy-1}{-2\left(x\sqrt{y}+\sqrt{xy}\right)}\)
\(R=\dfrac{\sqrt{x}+1}{x\sqrt{y}+\sqrt{xy}}\)
\(b.C=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(C=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(c.M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)
7 tháng 10 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(3+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{9-3}\)
\(\dfrac{x}{\sqrt{2}}=\dfrac{3+\sqrt{3}+3-\sqrt{3}}{6}=\dfrac{6}{6}=1\)
\(x=\sqrt{2}\)
\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
\(y\sqrt{2}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(y\sqrt{2}=\sqrt{7}+1-\sqrt{7}+1\)
\(y\sqrt{2}=2\)
\(y=\dfrac{2}{\sqrt{2}}\)
Thay \(x=\sqrt{2},y=\dfrac{2}{\sqrt{2}}\) vào A ta có:
\(A=\dfrac{\sqrt{2}.\dfrac{2}{\sqrt{2}}-1}{\sqrt{2}+\dfrac{2}{\sqrt{2}}}-\dfrac{1-\sqrt{2}.\dfrac{2}{\sqrt{2}}}{2\sqrt{2}-\dfrac{2}{\sqrt{2}}}\)
\(=\dfrac{2-1}{2\sqrt{2}}-\dfrac{1-2}{\sqrt{2}}\)
\(=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{4}\)
Tự kết luận nha