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Giải 2 bài luôn
Rút gọn:
\(Y=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{99}-\sqrt{100}}\)
\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+....+\sqrt{100}-\sqrt{99}\)
\(Y=\sqrt{10}-1\)
\(Y=9\)
Tính:
\(Y=\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+....+\frac{2014}{\sqrt{99}+\sqrt{100}}\)
\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(Y=\sqrt{10}-1\)
\(Y=9\)
\(Y=2014.9\)
\(Y=18126\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)
=\(-\sqrt{2}+\sqrt{2015}\)
Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014
\(A=\frac{\left(2\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\sqrt{5}+2\right)\left(\sqrt{5}+1\right)-\left(10+2\sqrt{5}\right)\left(\sqrt{5}-1\right)}{5-1}-1\)
\(=\frac{10+2\sqrt{5}+2\sqrt{5}+2-10\sqrt{5}+10-10+2\sqrt{5}}{4}-1\)
\(=\frac{12-4\sqrt{5}}{4}-1\)
\(=\frac{4\left(3-\sqrt{5}\right)}{4}-1\)
\(=3-\sqrt{5}-1\)
\(=2-\sqrt{5}\)
(còn biểu thức B hình như sai đề, bạn coi lại đề)
đề câu B nè : \(B=\sqrt{\left(1-\sqrt{2014}\right)^2}.\sqrt{2015+2\sqrt{2014}}\)
Ta xét biểu thức sau :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left[\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right]}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(với n > 0)
Áp dụng : \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+...+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)
A = \(\frac{1}{1+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + . . . . . . . . . + \(\frac{1}{\sqrt{99+\sqrt{100}}}\)
= \(\sqrt{2}\) - 1 + \(\sqrt{2}\) - \(\sqrt{3}\) + . . . . . . . + \(\sqrt{100}\) - \(\sqrt{99}\)
= - 1 + \(\sqrt{100}\) = \(\sqrt{100}\) - 1 = 10 - 1 = 9
\(A=\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2014}+\sqrt{2018}}\)
\(\Rightarrow A=\sqrt{5}-\sqrt{1}+\sqrt{9}-\sqrt{5}+...+\sqrt{2018}-\sqrt{2014}\)
\(\Rightarrow A=-\sqrt{1}+\sqrt{2018}\)
cho mk nha
Ai trên 11 điểm cho mình nha câu dưới 3 mk lại
\(\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+...+\frac{2014}{\sqrt{99}+\sqrt{100}}\)
\(=2014.\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)
\(=2014.\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)
\(=2014.\left(\sqrt{100}-\sqrt{1}\right)=2014.9=18126\)
\(\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+.....+\frac{2014}{\sqrt{9}+\sqrt{100}}\)
\(=\sqrt{1}-\sqrt{2}+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{999}\)
\(=\sqrt{100}-1\)
\(=9\)
P/s: Không chắc à