Tính tích
A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\cdot\dfrac{899}{900}\)
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21 tháng 6 2021
a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)
= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)
= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)
b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)
= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)
= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)
30 tháng 12 2020
[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4
= 1.1/3.(2.3)^8/(3.4)^4
= 1/3.2^8.3^8/3^4.4^4
= 1/3.2^8.3^8/3^4.2^8
= 1/3.3^8/3^4
= 1/3.3^4=27
(dấu . là nhân nha)
30 tháng 5 2022
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
TC
30 tháng 5 2022
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
18 tháng 3 2017
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
DN
1
15 tháng 4 2023
= 5/3 x ( 15/8 + 9/8) - 5/3
= 5/3 x 3 -5/3
=5- 5/3
= 10/3
nhớ tích cho mình nha mình cảm ơn rất nhìu
NS
1
NT
14 tháng 5 2018
A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\)=\(\dfrac{5}{3}\)=0.625
0.625 > 8998
like cho mình nha!
HT
16 tháng 3 2022
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
6 tháng 3 2018
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
18 tháng 2 2017
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{899}{900}\)
\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times...\times\frac{29.31}{30.30}\)
\(=\frac{\left(1\times2\times3\times...\times29\right)\left(3\times4\times5\times...\times31\right)}{\left(2\times3\times4\times...\times30\right)\left(2\times3\times4\times...\times30\right)}\)
\(=\frac{1\times2\times3\times...\times29}{2\times3\times4\times...\times30}.\frac{3\times4\times5\times...\times31}{2\times3\times4\times...\times30}\)
\(=\frac{1}{30}.\frac{31}{2}\)
\(=\frac{31}{60}\)
NT
19 tháng 2 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{29.31}{30.30}\\ =\frac{1.2.3.4....29}{2.3.4...30}.\frac{3.4.5...31}{2.3.4...30}\\ =\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
.
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\\ A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot29\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot31\right)}{\left(2\cdot3\cdot4\cdot...\cdot30\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot30\right)}\\ A=\dfrac{1\cdot2\cdot3\cdot...\cdot29}{2\cdot3\cdot4\cdot...\cdot30}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot31}{2\cdot3\cdot4\cdot...\cdot30}\\ A=\dfrac{1}{30}\cdot\dfrac{31}{2}\\ A=\dfrac{31}{60}\)
A=\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{899}{900}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
A=\(\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
A=\(\dfrac{\left(1.2.3.....28.29\right).\left(3.4.5....29.31\right)}{\left(2.3.4....29.30\right).\left(2.3.4....29.30\right)}\)
A=\(\dfrac{1.31}{30.2}\)
A=\(\dfrac{31}{60}\)
Vậy A = \(\dfrac{31}{60}\)