\(\frac{1}{2}\)+\(\frac{2}{3}\)=?
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25 tháng 8 2017
\(\frac{3}{1}+\frac{4}{5}=\frac{15}{5}+\frac{4}{5}=\frac{19}{5}\)
3 tháng 5 2017
H = 2012 - 1 - ( \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+99}\))
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{\left(99+1\right).\left[\left(99-1\right):1+1\right]:2}\)
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\))
= 2011 - 2.( \(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\))
= 2011 - 2.(\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))
= 2011 - 2.( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
= 2011 - 2.(\(\frac{1}{2}-\frac{1}{100}\)) = 2011 - 2.\(\frac{49}{100}\)= 2011 - \(\frac{49}{50}\)= \(\frac{100501}{50}\)
S
3 tháng 5 2017
\(H=2012-\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\right)\)
\(=2012-\left(1+\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{99\left(99+1\right):2}\right)\)
\(=2012-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2012-2\left(1-\frac{1}{100}\right)\)
\(=2012-2\cdot\frac{99}{100}\)
\(=2012-\frac{99}{50}\)
\(=\frac{100501}{50}\)
13 tháng 5 2018
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{2^2}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(2-\frac{1}{7}\right)=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\).
21 tháng 8 2019
\(\frac{5}{x}+\frac{4}{x+1}=\frac{3}{x+2}+\frac{2}{x+3}\)
\(\Leftrightarrow\frac{5\left(x+1\right)+4x}{x\left(x+1\right)}=\frac{3\left(x+3\right)+2\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{5x+5+4x}{x^2+x}=\frac{3x+9+2x+4}{x^2+5x+6}\)
\(\Leftrightarrow\frac{9x+5}{x^2+x}=\frac{5x+13}{x^2+5x+6}\)
\(\Leftrightarrow\left(9x+5\right)\left(x^2+5x+6\right)=\left(5x+13\right)\left(x^2+x\right)\)
\(\Leftrightarrow9x^3+45x^2+54x+5x^2+25x+30=5x^3+5x^2+13x^2+13x\)
\(\Leftrightarrow9x^3+50x^2+79x+30=5x^3+18x^2+13x\)
\(\Leftrightarrow9x^3-5x^3+50x^2-18x^2+79x-13x+30=0\)
\(\Leftrightarrow4x^3+32x^2+66x+30=0\)
\(\Leftrightarrow2x^3+16x^2+33x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2,3660\right)\left(x+0,6340\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x\approx2,3660\end{cases}or_{ }x\approx0,6340}\)
\(\frac{7}{6}nhé\)
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