Tính \(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
Tính gía trị biểu thức:
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2014\sqrt{2013}+2013\sqrt{2014}}+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
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
Rút gọn \(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}-...-\frac{1}{\sqrt{2013}+\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)
=\(-\sqrt{2}+\sqrt{2015}\)
So sánh 2 số:
\(a)\sqrt{2014}-\sqrt{2013};B=\sqrt{2015}-\sqrt{2014}\\ b)E=\frac{2014}{\sqrt{2015}}+\frac{2015}{\sqrt{2014}};F=\sqrt{2014}+\sqrt{2015}\)
Giải phương trình:
\(\frac{\sqrt{x-2013}-1}{x-2013}+\frac{\sqrt{y-2014}-1}{y-2014}+\frac{\sqrt{z-2015}-1}{z-2015}=\frac{3}{4}\)
Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)
\(\sqrt{y-2014}=b\left(b>0\right)\)
\(\sqrt{z-2015}=c\left(c>0\right)\)
Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)
<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)
<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)
<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).
Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)
\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)
\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)
=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)
Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
Tính giá trị biểu thức : \(S=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+.....+\frac{1}{2014\sqrt{2015}+2015\sqrt{2014}}\)
\(\sqrt{17}+\sqrt{26}+1và\sqrt{99}\)
b)chứng minh:\(\frac{1}{\sqrt{ }1}+\frac{1}{\sqrt{ }2}+\frac{1}{\sqrt{ }3}+...+\frac{1}{\sqrt{ }99}+\frac{1}{\sqrt{ }100}>10\)
c)cho:S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)vàP=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)tính \(\left(S-P\right)^{2016}\)
Nghiệm nhỏ nhất của pt\(\frac{1}{2\sqrt{x}-2014}+\frac{1}{3\sqrt{x}+2013}=\frac{1}{2015-4\sqrt{x}}+\frac{1}{9\sqrt{x}-2016}\)
Cho M=\(\frac{\sqrt{2}-\sqrt{1}}{1+1}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với 1/2