Tính :
\(M=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
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`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ P=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+...+\dfrac{1}{16}.\dfrac{16.17}{2}\\ P=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\\ P=\dfrac{1}{2}\left(2+3+4+...+17\right)\\ P=\dfrac{1}{2}\left(\dfrac{17.18}{2}-1\right)\\ P=\dfrac{1}{2}.152=76\)
\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\dfrac{1}{2}\left(2\cdot3:2\right)+\dfrac{1}{3}\cdot\left(3\cdot4:2\right)+...+\dfrac{1}{16}\left(16\cdot17:2\right)\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{2+3+4+...+17}{2}=\dfrac{152}{2}=76\)
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
\(M=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(M=1+\dfrac{1}{2}.3+\dfrac{1}{3}.6+...+\dfrac{1}{16}.136\)
\(M=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+...+\dfrac{1}{16}.\dfrac{16.17}{2}\)
\(M=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}=\dfrac{1}{2}\left(2+3+4+...+17\right)\)
\(M=\dfrac{1}{2}.\left(\dfrac{17.18}{2}-1\right)=\dfrac{1}{2}.152=76\)