TÍNH
\(\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot\left(1000-3^3\right)\cdot\cdot\cdot\left(1000-50^3\right)\)
Những câu hỏi liên quan
Tính A = \(\left(1^3-1000\right)\cdot\left(2^3-1000\right)\cdot...\cdot\left(2018^3-1000\right)\).Kb vs mik nha! ...
\(A=\)\(\left(1^3-1000\right).\left(2^3-1000\right)\)\(.....\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...\left(10^3-1000\right)...\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...0...\left(2018^3-1000\right)\)
\(A=0\)
~~~Hok tốt~~~
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A)2009^{left(1000-1^3right)cdotleft(1000-2^3right)cdot...cdotleft(1000-15^3right)}B)left(frac{1}{125}-frac{1}{1^3}right)cdotleft(frac{1}{125}-frac{1}{2^3}right)cdot...cdotleft(frac{1}{125}-frac{1}{25^3}right)C)left(frac{1}{38}-1right)cdotleft(frac{1}{37}-1right)cdotleft(frac{1}{36}-1right)cdot...cdotleft(frac{1}{2}-1right)HELP ME!!!!!!!!!!!!!!!!!!!
Đọc tiếp
A)\(2009^{\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-15^3\right)}\)
B)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
C)\(\left(\frac{1}{38}-1\right)\cdot\left(\frac{1}{37}-1\right)\cdot\left(\frac{1}{36}-1\right)\cdot...\cdot\left(\frac{1}{2}-1\right)\)
HELP ME!!!!!!!!!!!!!!!!!!!
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
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\(2009^{\left(1000-1^3\right)\cdot\left(100-2^3\right)\cdot...\left(1000-15^3\right)}\)
mik dang can gap
\(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-10^3\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-1000\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...0..\left(1000-15^3\right)}\)
= \(2009^0\)
= \(1\)
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Tính nhanh :\(\left(3-\frac{1}{18}\cdot\right)\cdot\left(3-\frac{2}{18}\right)\cdot\left(3-\frac{3}{18}\right)\cdot\cdot\cdot\cdot\left(3-\frac{2019}{18}\right)\)
Tính:-3^2+left{-54:left[-2^8+7right]cdotleft(-2right)^2right}Tính hợp lí :31cdotleft(-18right)+31cdotleft(-81right)-31left(-12right)cdot47+left(-12right)cdot52+left(-12right)13cdotleft(23+22right)-3cdotleft(17+28right)-48+48cdotleft(-78right)+48cdotleft(-21right)
Đọc tiếp
Tính:
\(-3^2+\left\{-54:\left[-2^8+7\right]\cdot\left(-2\right)^2\right\}\)
Tính hợp lí :
\(31\cdot\left(-18\right)+31\cdot\left(-81\right)-31\)
\(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)
\(13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\)
\(-48+48\cdot\left(-78\right)+48\cdot\left(-21\right)\)
`#3107.101107`
`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`
`= -9 + [-54 \div (-256 + 7) * 4]`
`= -9 + [-54 \div (-249) * 4]`
`= -9 + (18/83 * 4)`
`= -9 + 72/83`
`= -675/83`
______
`31 * (-18) + 31 * (-81) - 31`
`= 31 * (-18 - 81 - 1)`
`= 31 * (-100)`
`= -3100`
___
`(-12) * 47 + (-12) * 52 + (-12)`
`= (-12) * (47 + 52 + 1)`
`= (-12) * 100`
`= -1200`
___
`13 * (23 + 22) - 3 * (17 + 28)`
`= 13 * 45 - 3 * 45`
`= 45 * (13 - 3)`
`= 45 * 10`
`= 450`
____
`-48 + 48 * (-78) + 48 * (-21)`
`= 48 * (-1 - 78 - 21)`
`= 48 * (-100)`
`= -4800`
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TÍNH
\(C=\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{2}{2015}\right)\cdot\left(1+\frac{2}{2017}\right)\)
\(D=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\left(1-\frac{1}{10}\right)\cdot\left(1-\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
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tính \(a=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\cdot\cdot\cdot+\frac{1}{32}\cdot\left(1+2+3+\cdot\cdot\cdot+32\right)\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)
\(\frac{2+3+4+....+33}{2}\)
\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)
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G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
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\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
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tính:
D=\(\left[\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{25}\right)\right]\):\(\left[\left(1+\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{25}\right)\right]\)