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E hổng biết cách này có đúng ko nữa:((
5
Ta có:\(S=\frac{2010}{x}+\frac{1}{2010y}+\frac{1010}{1005}\ge2\sqrt{\frac{2010}{x}\cdot\frac{1}{2010y}}+\frac{1010}{1005}\left(AM-GM\right)\)
\(=\frac{2}{\sqrt{xy}}+\frac{2010}{1005}\ge\frac{2}{\frac{x+y}{2}}+2=4\)( AM-GM ngược dấu )
Dấu "=" xảy ra khi \(x=y=\frac{2010}{4024}\)
\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)
\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
Làm nốt
Ta có:
\(\left[x+\sqrt{\left(x+2010\right)}\right].\left[\sqrt{\left(x+2010\right)-x}\right]=2010\)
\(\Rightarrow\sqrt{\left(x-2010\right)-x}=\sqrt{\left(x+2010\right)+y}\left(1\right)\)
\(\Leftrightarrow\sqrt{\left(y+2010\right)-y}=\sqrt{\left(x+2010\right)+x}\left(2\right)\)
Công 2 vé lại với nhau, ta có:
\(\Rightarrow\sqrt{\left(x+2010\right)}+\sqrt{\left(y+2010\right)}-x-y=\sqrt{\left(x+2010\right)}+\sqrt{\left(y+2010\right)}+x+y\)
\(\Leftrightarrow2\left(x+y\right)=0\)
\(\Rightarrow x^3+y^3=0\)
(x-√(x^2+2010).(x+√(x^2+2010)).(y+√(y^2+... = 2010.(x-√(x^2+2010)
<=> -2010.(y+√(y^2+2010) = 2010.(x-√(x^2+2010)
<=> - (y+√(y^2+2010) = (x-√(x^2+2010)
<=> (x-√(x^2+2010) = - (y+√(y^2+2010)
+++ (x+√(x^2+2010)) (y+√(y^2+2010))(y-√(y^2+2010)) = 2010.(y-√(y^2+2010))
<=> -2010.(x+√(x^2+2010) = 2010.(y-√(y^2+2010))
<=> - (x+√(x^2+2010) = (y-√(y^2+2010) (**)
...Lấy (*) - (**) vế theo vế,ta có:
2x = -2y
<=> x + y = 0
\(\left[\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}\right]\left[\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]^2=\left(x+\sqrt{x}+1\right)\frac{1}{\left(1+\sqrt{x}\right)^2}=\frac{x+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
Đề bài sai
\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(\sqrt{2011}-\sqrt{2010}=\frac{1}{\sqrt{2011}+\sqrt{2010}}\)
Do \(\sqrt{2012}>\sqrt{2010}\) \(\Rightarrow\sqrt{2012}+\sqrt{2011}>\sqrt{2011}+\sqrt{2010}>0\)
\(\Rightarrow\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\Rightarrow\sqrt{2012}-\sqrt{2011}< \sqrt{2011}-\sqrt{2010}\)
\(A=\frac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
\(M^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\le2\left(x-1+9-x\right)=16\)
\(\Rightarrow M\le4\Rightarrow M_{max}=4\) khi \(x-1=9-x\Leftrightarrow x=5\)
đặt \(\sqrt{x+2}\)=a ; \(\sqrt{2-x}\)=b
=>\(\frac{a+b}{a-b}=\sqrt{2}\)<=> a+b = a\(\sqrt{2}-b\sqrt{2}\)<=> a(\(\sqrt{2}-1\))= b(\(1+\sqrt{2}\))
<=> \(\frac{a}{b}=\frac{\sqrt{2}+1}{\sqrt{2}-1}=\left(\sqrt{2}+1\right)^2\)
=> \(\frac{x+2}{2-x}=\frac{a^2}{b^2}=\left(\sqrt{2}+1\right)^4\)
\(\left(2n+1\right)^2=4n^2+4n+1\)
\(>4n^2+4n=4n\left(n+1\right)\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
\(\Rightarrow\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{1}{2}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\) \(=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{1}{2}\cdot\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\right)\)
\(< \frac{1}{2}\)