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Đặt \(\hept{\begin{cases}\sqrt{1+\frac{\sqrt{3}}{2}}=a\\\sqrt{1-\frac{\sqrt{3}}{2}}=b\end{cases}}\)
\(\Rightarrow a^2+b^2=2;ab=\frac{1}{2};a-b=1\)
\(\Rightarrow\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{a^2}{1+a}+\frac{b^2}{1-b}\)
\(=\frac{a^2+b^2-ab\left(a-b\right)}{1-ab+\left(a-b\right)}=\frac{2-\frac{1}{2}.1}{1-\frac{1}{2}+1}=1\)
\(\frac{4.\left(\sqrt{3}+1\right)}{\sqrt{3}-1}-\frac{2+\sqrt{3}}{2-\sqrt{3}}\)
\(\Leftrightarrow\frac{4\left(\sqrt{3}+1\right)\left(2-\sqrt{3}\right)}{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}-\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(2-\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
\(\Leftrightarrow\frac{4\left(\sqrt{3}+1\right)\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}\)
\(\Rightarrow\frac{3\sqrt{3}-5}{3\sqrt{5}-5}=1\left(đpcm\right)\)
a, \(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(\Rightarrow\) \(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(\Rightarrow\) \(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\Rightarrow\) \(2S=1-\frac{1}{2017}\)
\(\Rightarrow\) \(2S=\frac{2016}{2017}\)
\(\Rightarrow\) \(S=\frac{1008}{2017}\)
\(\frac{1}{2\sqrt{n+1}}=\frac{1}{\sqrt{n+1}+\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
=> \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)(1)
\(\frac{1}{2\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)=> \(\frac{1}{2\sqrt{n}}>\sqrt{n+1}-\sqrt{n}\)(2)
Từ (1) và (2) => \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}< \frac{1}{2\sqrt{n}}\)
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}=1+\frac{1}{n}-\frac{1}{n+1}\)
\(S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+....+1+\frac{1}{n}-\frac{1}{n+1}\)
\(=n+1-\frac{1}{n+1}=\frac{\left(n+1\right)^2-1}{n+1}=\frac{2009^2-1}{2009}\Rightarrow n+1=2009\Rightarrow n=2008\)
a) ĐK: a>0, a khác 1, a khác 1/4
P=\(1+\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{a\sqrt{a}}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}\right).\frac{\sqrt{a}-1}{2\sqrt{a}-1}=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a\left(\sqrt{a}-1\right)}-\frac{\left(2\sqrt{a}-1\right)a}{a\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}-1}{2\sqrt{a}-1}\)
\(P=\frac{2a\sqrt{a}-a-2\sqrt{a}+1-2a\sqrt{a}+a}{a\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}-1}{2\sqrt{a}-1}=\frac{-\left(2\sqrt{a}-1\right)}{a\left(2\sqrt{a}-1\right)}=-\frac{1}{a}\)
b)
ta có: a>0 => -1/a<0 ; 2/3>0 => Pkhông thể > 2/3 đc. bạn xem lại đề rồi có gì liên hệ vs mình nha.
nhớ L IK E