A=1-\(\dfrac{3}{4}\)-\(\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{4}\right)^3+\left(\dfrac{3}{4}\right)^4-...+\left(\dfrac{3}{4}\right)^{2023}\)
CM :A ko là số nguyên
K
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14 tháng 8 2021
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
6 tháng 9 2021
A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))3
A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)
A = \(\dfrac{-73}{50}\)
6 tháng 9 2021
B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))
B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)
B = \(\dfrac{13}{12}\)
AH
Akai Haruma
Giáo viên
10 tháng 10 2023
Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$
$=\frac{147}{64}+1=\frac{211}{64}$
c.
$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$
$=-4(\frac{-7}{4})^2=\frac{-49}{4}$
KV
11 tháng 10 2023
a) = 0,16 - 0,064 . (-3)
= 0,16 + 0,192
= 0,352
b) = (7/4)³ - (7/4)² + 1
= 343/64 - 49/16 + 1
= 147/64 + 1
= 211/64
c) = 8/27 - 4.(-7/4)² - 8/27
= -4.49/16
= -49/4
29 tháng 8 2017
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
13 tháng 6 2023
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
20 tháng 12 2021
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
26 tháng 2 2023
\(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{2023}\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.\dfrac{6}{5}.\dfrac{7}{6}.\dfrac{8}{7}...\dfrac{2021}{2020}.\dfrac{2022}{2021}.\dfrac{2023}{2022}.\dfrac{2024}{2023}\)
\(=\dfrac{2024}{2}=1012\)