tính nhanh
B = \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(C=\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\cdot\dfrac{3-2-1}{6}=0\)
a, \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
= \(\dfrac{28}{15}.\dfrac{25}{100}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{32-79}{60}\right).\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-47}{60}.\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-24}{60}=\dfrac{60}{60}=1\)
b, \(\dfrac{\left(\dfrac{11^2}{200}+0,415\right):0,01}{\dfrac{1}{12}-37,25+3\dfrac{1}{6}}\)
= \(\dfrac{\left(\dfrac{121}{200}+\dfrac{415}{1000}\right):\dfrac{1}{100}}{\dfrac{1}{12}-\dfrac{3725}{100}+\dfrac{19}{6}}=\dfrac{\left(\dfrac{121}{200}+\dfrac{83}{200}\right).100}{\dfrac{1}{12}-\dfrac{149}{4}+\dfrac{19}{6}}\)
= \(\dfrac{\dfrac{51}{50}.100}{-34}=\dfrac{102}{-34}=-3\)
a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)
\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
câu dễ như này sao bạn không tự giải ? cứ phụ thuộc vào người khác thôi
\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)
\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)
\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)
\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)
\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)
Câu 2 mik chịu r sorry:(
Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.
Bai 1: tính nhanh A) -5/9 + 3/5 - 3/9 + -2/5 B) -5/13 + (3/5 + 3/1 - 4/10) C) 5/17 - 9/15 - 2/-17 + -2/15 D) (1/9 - 9/17) + 3/6 - ( 12/17 - 1/2) + -1/9 Bài 5: tính tổng A) 1/3 + -1/4 + 1/5 + 1/-6 + -1/-7 + 1/6 + -1/5 + 1/4 + 1/3 B) 1/12 +1/2.3+1/3.4+..+1/99100 Giúp mình nhé nhanh
c: Ta có: \(-\dfrac{5}{13}-\left(\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{4}{10}\right)\)
\(=\dfrac{-5}{13}-\dfrac{3}{5}-\dfrac{3}{13}+\dfrac{2}{5}\)
\(=\dfrac{-8}{13}-\dfrac{1}{5}\)
\(=\dfrac{-53}{65}\)
d: Ta có: \(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{5}{9}\)
\(=\dfrac{1}{9}-\dfrac{9}{17}+\dfrac{1}{2}-\dfrac{12}{17}+\dfrac{1}{2}+\dfrac{5}{9}\)
\(=\dfrac{2}{3}+1-\dfrac{21}{17}\)
\(=\dfrac{22}{51}\)
`(12/999 +23/200 -34/201 ) .(1/2-1/3-1/6)`
`= (12/999 +23/200 -34/201 ) .( 3/6 - 2/6-1/6)`
` =(12/999 +23/200 -34/201 ) . 0/6`
` =(12/999 +23/200 -34/201 ) . 0`
`=0`