\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\righ...">
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2 tháng 11 2019

\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+...+\frac{1}{16}.\frac{16.\left(16+1\right)}{2}\)

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)

\(=76\)

2 tháng 11 2019

\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

     \(=1+\frac{1}{2}\left[\frac{\left(2+1\right)2}{2}\right]+\frac{1}{3}\left[\frac{\left(3+1\right)3}{3}\right]+...+\frac{1}{16}\left[\frac{\left(16+1\right)16}{2}\right]\)

      \(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)

        \(=\frac{2+2+1+3+1+...+16+1}{2}\)

          \(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)

            \(=\frac{15+135+2}{2}\)

              \(=\frac{152}{2}\)\(=76\)

7 tháng 2 2017

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(=\frac{\frac{17.18}{2}-1}{2}=76\)

a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)

b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)

c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)

15 tháng 3 2020

Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)

=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)

\(=\frac{5}{2}.\frac{6}{5}-17\)

\(=3-17=-14\)

Tụi quá mới lớp 5 thui

\(2^3+3.\left(\frac{2}{3}\right)^0-2+\left[\left(-2\right)^2:\frac{1}{2}\right]-8\)

đổi p/s \(\left(\frac{2}{3}\right)^0=1\)

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