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\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
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
\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)
b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{2\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}=\dfrac{6\sqrt{3}+2}{13}\)
\(P=\left(\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}\right):\left(\dfrac{x+y+2xy}{1-xy}+1\right)\)
Điều kiện : \(xy\ge0\) hoặc \(xy\le0\) ; \(xy\ne1\); \(x\ge0\);\(y\ge0\)
\(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\right):\left(\dfrac{x+2xy+y+1-xy}{1-xy}\right)\)
\(P=\left(\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}\right):\left(\dfrac{x+xy+y+1}{1-xy}\right)\)
\(P=\left(\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\dfrac{x\left(1+y\right)+\left(y+1\right)}{1-xy}\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}\right):\left(\dfrac{\left(1+y\right)\left(x+1\right)}{1-xy}\right)\)
\(P=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+y\right)\left(x+1\right)}\)
\(P=\dfrac{2\sqrt{x}}{x+1}\)
b) ta có :\(x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{4-3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
thay \(x=\left(\sqrt{3}-1\right)^2\) vào biểu thức P
ta được : \(P=\dfrac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+1}\)
\(P=\dfrac{2\left|\sqrt{3}-1\right|}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}\)
\(P=\dfrac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\dfrac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)
\(P=\dfrac{6\sqrt{3}+2}{13}\)
c) để P\(\le\)1 thì \(\dfrac{2\sqrt{x}}{x+1}\le1\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x+1}-1\le0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-x-1}{x+1}\le0\)
\(\Leftrightarrow\dfrac{-\left(x-2\sqrt{x}+1\right)}{x+1}\le0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x+1}\le0\)
Vì \(-\left(x-1\right)^2\le0\) nên x + 1 \(\ge\) 0
\(\Leftrightarrow\) x \(\ge\) -1
đúng thì cho xin 1 like nha
ĐKXĐ : \(x,y>0\)
a) \(A=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right).\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
\(A=\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)
\(A=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\right):\dfrac{\sqrt{x}\left(x+y\right)+\sqrt{y}\left(x+y\right)}{\sqrt{xy}\left(x+y\right)}\)
\(A=\dfrac{2\sqrt{xy}+x+y}{xy}.\dfrac{\sqrt{xy}\left(x+y\right)}{\sqrt{x}\left(x+y\right)+\sqrt{y}\left(x+y\right)}\)
\(A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{xy}}.\dfrac{x+y}{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(A=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
b) \(A=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\ge\dfrac{2\sqrt[4]{xy}}{\sqrt{xy}}=\dfrac{2\sqrt[4]{16}}{\sqrt{16}}=1\) ( Cosi )
Vậy GTNN của A là \(1\) khi \(x=y=4\)
Chúc bạn học tốt ~