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NT
1
24 tháng 9 2017
ĐKXĐ: x;y >=0
Ta có : \(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)
\(\Leftrightarrow x+y+13=4\sqrt{x}+6\sqrt{y}\)
\(\Leftrightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)
\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-6\sqrt{y}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y}-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=9\end{cases}}\left(TMĐKXĐ\right)}\)
29 tháng 12 2021
d: \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=4\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
10 tháng 8 2017
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
30 tháng 6 2018
\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)
\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)
\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)
TT
1 tháng 7 2018
Câu a
\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
DT
4
\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)
\(\Leftrightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)
\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)
Xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}=2\\\sqrt{y}=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)
Điều kiện xác định: \(x,y>0\)
\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)
\(\Leftrightarrow x-4\sqrt{x}+4+y-6\sqrt{y}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y}-3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2=0}{\sqrt{y}-3=0}\)
\(\Leftrightarrow\dfrac{x=4}{y=9}\)
KL: x = 4, y = 9