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1. Ta có : \(A=\frac{\left(x+4\right)\left(x+9\right)}{x}=\frac{x^2+13x+36}{x}=x+\frac{36}{x}+13\)
Áp dụng bđt Cauchy : \(x+\frac{36}{x}\ge2\sqrt{x.\frac{36}{x}}=12\)
\(\Rightarrow A\ge25\)
Vậy Min A = 25 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{36}{x}\end{cases}\) \(\Leftrightarrow x=6\)
2. \(B=\frac{\left(x+100\right)^2}{x}=\frac{x^2+200x+100^2}{x}=x+\frac{100^2}{x}+200\)
Áp dụng bđt Cauchy : \(x+\frac{100^2}{x}\ge2\sqrt{x.\frac{100^2}{x}}=200\)
\(\Rightarrow B\ge400\)
Vậy Min B = 400 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{100^2}{x}\end{cases}\) \(\Leftrightarrow x=100\)
A=\(1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
A= \(\left(x+\dfrac{1}{2x}\right)+\left(y+\dfrac{1}{2y}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+2\)
Áp Dụng BĐT Cô si ta có:
\(\left(x+\dfrac{1}{2x}\right)\ge\sqrt{2}\); \(\left(y+\dfrac{1}{2y}\right)\ge\sqrt{2}\); \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)
\(\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{\dfrac{1}{2x.2y}}=\dfrac{1}{\sqrt{xy}}\ge\dfrac{\sqrt{2}}{\sqrt{x^2+y^2}}=\sqrt{2}\)
suy ra A\(\ge4+3\sqrt{2}\)
Dấu = xảy ra
\(\left\{{}\begin{matrix}x=y\\x=\dfrac{1}{2x}\\y=\dfrac{1}{2y}\end{matrix}\right.\)
\(\Leftrightarrow\)x=y=\(\dfrac{\sqrt{2}}{2}\)
Vậy Min A=4+3\(\sqrt{2}\) khi x=y=\(\dfrac{\sqrt{2}}{2}\)
Trước hết ta có \(\dfrac{\left(x+y\right)^2}{2}\le x^2+y^2\Rightarrow x+y\le\sqrt{2\left(x^2+y^2\right)}=\sqrt{2}\)
\(A=1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
\(A=2+x+y+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{x}{y}+\dfrac{y}{x}\ge2+x+y+\dfrac{4}{x+y}+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)
\(\Rightarrow A\ge4+x+y+\dfrac{4}{x+y}=4+x+y+\dfrac{2}{x+y}+\dfrac{2}{x+y}\)
\(\Rightarrow A\ge4+2\sqrt{\left(x+y\right).\dfrac{2}{\left(x+y\right)}}+\dfrac{2}{\sqrt{2}}=4+3\sqrt{2}\)
\(\Rightarrow A_{min}=4+3\sqrt{2}\) khi \(x=y=\dfrac{1}{\sqrt{2}}\)
Bài 1:
a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
b: \(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{2xy}+\frac{\left(x+y\right)^2}{2xy}\)
\(S\ge\frac{4\left(x+y\right)^2}{x^2+y^2+2xy}+\frac{\left(x+y\right)^2}{\frac{\left(x+y\right)^2}{2}}=\frac{4\left(x+y\right)^2}{\left(x+y\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\) khi \(x=y\)
a,\(A\ge\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\ge\frac{9}{\sqrt{3\left(x+y+z\right)}}=3\)=3
MInA=3<=>x=y=z=1
b)dùng cô si đi(đề thi chuyên bình phước năm 2016-2017)
Áp dụng: \(\left(a+b\right)^2\ge4ab\)
\(B=\dfrac{\left(x+2020\right)^2}{x}\ge\dfrac{4.2020.x}{x}=8080\)
\(B_{min}=8080\) khi \(x=2020\)