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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 \(x=y=4\)
2
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
a) \(P=\dfrac{\left(x^2+2xy+9y^2\right)-\left(x+3y-2\sqrt{xy}\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x^2+6xy+9y^2\right)-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)^2-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)\left(x+3y-2\sqrt{xy}\right)}{x+3y-2\sqrt{xy}}\)
\(P=x+3y\)
b) \(\dfrac{P}{\sqrt{xy}+y}=\dfrac{x+3y}{\sqrt{xy}+y}=\dfrac{\left(x+3y\right):y}{\left(\sqrt{xy}+y\right):y}=\dfrac{\dfrac{x}{y}+3}{\sqrt{\dfrac{x}{y}}+1}\)
Đặt \(t=\sqrt{\dfrac{x}{y}}>0\) và \(\dfrac{P}{\sqrt{xy}+y}=Q\) thì \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)
\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)
\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)
\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)
\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)
\(=5\left(a+b\right)=5.2016=10080\)
áp dụng bdt cauchy -schửat dạng engel ta có
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\)
(do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé)
dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
\(M=x^2-\sqrt{x^3y}-\sqrt{xy^3}+y^2\)
\(=\sqrt{x^3}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{y^3}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x^3}-\sqrt{y^3}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2\left(x+\sqrt{xy}+y\right)\ge0\) \(\forall x;y\ge0\)