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Ta có
\(\hept{\begin{cases}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\left(1\right)\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\left(2\right)\end{cases}}\)
\(\Rightarrow\frac{1}{\sqrt{2008}+\sqrt{2005}}+\frac{1}{\sqrt{2010}+\sqrt{2007}}>\frac{2}{\sqrt{2015}+\sqrt{2009}}\)
\(\Leftrightarrow\frac{\sqrt{2008}-\sqrt{2005}}{3}+\frac{\sqrt{2010}-\sqrt{2007}}{3}>\frac{\sqrt{2015}-\sqrt{2009}}{3}\)
\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)
\(A-B=\sqrt{2009}-\sqrt{2007}+\sqrt{2010}-\sqrt{2008}+\sqrt{2011}-\sqrt{2015}\)
\(=\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}\)
Ta có \(\left\{{}\begin{matrix}\sqrt{2009}+\sqrt{2007}< \sqrt{2011}+\sqrt{2015}\\\sqrt{2010}+\sqrt{2008}< \sqrt{2011}+\sqrt{2015}\end{matrix}\right.\)
\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}>\frac{2}{\sqrt{2011}+\sqrt{2015}}+\frac{2}{\sqrt{2011}+\sqrt{2015}}=\frac{4}{\sqrt{2011}+\sqrt{2015}}\)
\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}>0\)
\(\Rightarrow A-B>0\Rightarrow A>B\)
a) A = \(\sqrt{3}-\sqrt{2}\) và B = \(\sqrt{6}-\sqrt{5}\)
Giả sử : \(\sqrt{3}-\sqrt{2}\le\sqrt{6}-\sqrt{5}\)
⇔ \(\left(\sqrt{3}-\sqrt{2}\right)^2\le\left(\sqrt{6}-\sqrt{5}\right)^2\)
⇔ 5 - \(2\sqrt{6}\) ≤ 11 - \(2\sqrt{30}\)
⇔ \(2\sqrt{30}-2\sqrt{6}\) ≤ 6
⇔\(\left(2\sqrt{30}-2\sqrt{6}\right)^2\le6^2\)
⇔ 36, 66 ≤ 36 (sai)
Vậy \(\sqrt{3}-\sqrt{2}>\sqrt{6}-\sqrt{5}\)
a) ta có : \(8+2\sqrt{15}>8+2\sqrt{12}\Leftrightarrow\left(\sqrt{3}+\sqrt{5}\right)^2>\left(\sqrt{6}+\sqrt{2}\right)^2\)
\(\Leftrightarrow\sqrt{3}+\sqrt{5}>\sqrt{6}+\sqrt{2}\Leftrightarrow\sqrt{3}-\sqrt{2}>\sqrt{6}-\sqrt{5}\)
câu b tương tự nha
a: \(A=\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(B=\dfrac{1}{\sqrt{6}+\sqrt{5}}\)
mà 3<6; 2<5
nên A>B
a: \(A=\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(B=\dfrac{1}{\sqrt{6}+\sqrt{5}}\)
mà 3<6; 2<5
nên A>B
b: A=134,6327
B=134,6328
Do đó: A<B
a. Ta có \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\to\frac{1}{\sqrt{2016}+\sqrt{2015}}<\frac{1}{\sqrt{2015}+\sqrt{2014}}\). Nhân liên hợp từng phân thức, ta có
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}<\frac{\sqrt{2015}-\sqrt{2014}}{\left(\sqrt{2015}+\sqrt{2014}\right)\left(\sqrt{2015}-\sqrt{2014}\right)}\)
\(\Leftrightarrow\sqrt{2016}-\sqrt{2015}<\sqrt{2015}-\sqrt{2014}\Leftrightarrow\sqrt{2016}+\sqrt{2014}<2\sqrt{2015}.\)
b. Tiếp tục thực hiện các biến đổi liên hợp, ta có
\(\sqrt{2008}-\sqrt{2005}+\sqrt{2009}-\sqrt{2007}=\frac{3}{\sqrt{2008}+\sqrt{2005}}+\frac{2}{\sqrt{2009}+\sqrt{2007}}\)
\(>\frac{3}{\sqrt{2015}+\sqrt{2010}}+\frac{2}{\sqrt{2015}+\sqrt{2010}}=\frac{5}{\sqrt{2015}+\sqrt{2010}}=\sqrt{2015}-\sqrt{2010}\)
Suy ra \(\sqrt{2008}-\sqrt{2005}+\sqrt{2009}-\sqrt{2007}>\sqrt{2015}-\sqrt{2010}\to\)
\(\to\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}.\) (ĐPCM).