Tính giá trị
S=\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+100}\right)\)
Tính giá trị biểu thức sau:
\(D=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
tính : \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
Ta thấy \(1-\dfrac{1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\) với mọi \(n>0\).
Từ đó \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}...\dfrac{99.101}{100}=\left(\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{99}{100}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{101}{100}\right)=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\).
Tính: \(B=\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\) rồi so sánh với \(\dfrac{1}{2}\)
\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(B=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)\cdot\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)....\left(\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\right)\)
\(B=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}....\cdot\dfrac{100^2-1}{100^2}\)
\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot...\cdot\dfrac{\left(100+1\right)\left(100-1\right)}{100^2}\)
\(B=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}\)
\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot101}{2^2\cdot3^2\cdot4^2\cdot5^2\cdot....\cdot100^2}\)
\(B=\dfrac{1\cdot101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
\(B=\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
Mà: \(\dfrac{1}{2}=\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
Ta có: \(101< 3\cdot4\cdot5\cdot...\cdot100\)
\(\Rightarrow\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}< \dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
\(\Rightarrow B< \dfrac{1}{2}\)
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
Tính A= \(\dfrac{1}{2}+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+...+\dfrac{1}{100}.\left(1+2+3+...+100\right)\)
Áp dụng \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{n}\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2n}=\dfrac{n+1}{2}\)
Vậy:
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}=\dfrac{1+2+3+...+100}{2}-1\)
\(=\dfrac{100.101}{2}-1=5049\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
`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`
Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
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\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)