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a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\x\ne y\end{matrix}\right.\)
Gọi biểu thức trên là A , ta có:
\(A=\frac{2\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\frac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{3\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\frac{2\sqrt{x}-2\sqrt{y}+\sqrt{x}+\sqrt{y}-3\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\frac{-\sqrt{y}}{x-y}\left(=\frac{\sqrt{y}}{y-x}\right)\)
b) Với x=4 ; y=9 ta có:
\(A=\frac{\sqrt{9}}{9-4}=\frac{3}{5}\)
c) Ta có: với x>y>0 thì A<=>\(\left\{{}\begin{matrix}\sqrt{y}>0\\x>y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}>0\\y-x< 0\end{matrix}\right.\Leftrightarrow A< 0\)
Vậy A<0 với mọi x>y>0
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
\(A=\left(\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{y}-\sqrt{x}}\right):\dfrac{2\sqrt{xy}}{x-y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)
b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)
Vậy pt vô nghiệm
chịu thua vô điều kiện xin lỗi nha : v
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
a)\(B=\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{0}+\sqrt{4}}=\frac{1}{2}\)
b)\(M=A+B=\frac{2\sqrt{y}}{x-y}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)\(=\frac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{2\sqrt{y}+\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)\(=\frac{2\sqrt{y}+2\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2}{\sqrt{x}-\sqrt{y}}\)
c)\(M=\frac{2}{\sqrt{x}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{4y}-\sqrt{y}}\)<=>\(1=\frac{2}{2\sqrt{y}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{y}}\)<=> \(\sqrt{y}=2\)
<=> \(\left(\sqrt{y}\right)^2=2^2\)<=> \(y=4\)
=>\(x=4y=4\cdot4=16\)