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1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Đặt A=\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
A=\(\dfrac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}\)
A=\(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
A=2.\(5^2\)
A=2.25
A=50
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
\(C=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2^2\right)\\ C=\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\dfrac{-4}{1}\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\\ C=\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\\ C=\dfrac{97}{56}\)
`8/5 .2/3 + (-5.5)/(3.5) = 16/15 - 5/3 = -3/5`
b) 6/7+5/8 :5 -3/16 .(-2)^2=6/7 + 1/8 - 3/16 .4`
`=55/56 - 3/4`
`=13/56`
\(a,\dfrac{8}{5}.\dfrac{2}{3}+\dfrac{-5.5}{3.5}=\dfrac{16}{15}+\dfrac{-25}{15}=-\dfrac{9}{15}=-\dfrac{3}{5}\)
\(b,\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{55}{56}-\dfrac{3}{4}=\dfrac{13}{56}\)