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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.\(DK:x,y>0\)
Ta co:
\(A=\frac{x+y+2\sqrt{xy}}{xy}.\frac{\sqrt{xy}\left(x+y\right)}{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
b.
Ta lai co:
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\ge\frac{2\sqrt{\sqrt{x}.\sqrt{y}}}{4}=1\)
Dau '=' xay ra khi \(x=y=4\)
Vay \(A_{min}=1\)khi \(x=y=4\)
ĐKXĐ:
\(P=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x+y}{xy}\right]:\left[\frac{\sqrt{x}\left(x+y\right)+\sqrt{y}\left(x+y\right)}{\sqrt{xy}\left(x+y\right)}\right]\)
\(=\left(\frac{2\sqrt{xy}+x+y}{xy}\right):\left[\frac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\right]=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}.\frac{\sqrt{xy}}{\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
\(xy=16\Rightarrow\left\{{}\begin{matrix}\sqrt{xy}=4\\y=\frac{16}{x}\end{matrix}\right.\)
\(\Rightarrow P=\frac{\sqrt{x}+\frac{4}{\sqrt{x}}}{4}\ge\frac{1}{4}\left(2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}\right)=1\)
\(\Rightarrow P_{min}=1\) khi \(x=y=4\)
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 = \(\sqrt{6}\)
