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23 tháng 7 2017
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
23 tháng 7 2017
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
T
23 tháng 8 2023
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
Q
9 tháng 7 2017
3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
NH
11 tháng 3 2023
\(A=\dfrac{4}{1\cdot2}+\dfrac{4}{2\cdot3}+\dfrac{4}{3\cdot4}+...+\dfrac{4}{2022\cdot2023}\\ A=4\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\\ A=4\cdot\dfrac{2022}{2023}\\ A=\dfrac{8088}{2023}\)
NV
7 tháng 7 2018
\(c,1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
\(d,\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=2\)
MD
12 tháng 3 2017
Câu 1 : \(1,321338308x10^{-4}\)
Câu 2 : \(1316,572106\)
Câu 3 : \(1,641302619x10^{-13}\)
Ủng hộ nhé,tớ đang âm.
TB
3
1 tháng 8 2023
\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)
KS
2 tháng 8 2023
`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`
`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`
`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`
`= 1/2 - 1/4090506`
`=4090506/8181012 - 2/8181012`
`= 4090504/8181012`
DD
2
MV
10 tháng 7 2017
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
\(-\dfrac{3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}\)
\(=-\dfrac{3^4\cdot2^8}{2^4\cdot3^2}\)
\(=-\dfrac{3^4\cdot2^8}{2^4\cdot3^2}\)
\(=-\dfrac{3^2\cdot2^4}{1\cdot1}\)
\(=-9\cdot16\)
\(=-144\)
`@` `\text {Ans}`
`\downarrow`
\(-\dfrac{3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}\)
`=`\(-\dfrac{3^4\cdot2^8}{2^4\cdot3^2}=-3^2\cdot2^4=-\left(12\right)^2=-144\)