Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)
\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)
\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x\left(1-\dfrac{1}{34}\right)=33\)
\(\dfrac{33}{34}x=33\)
\(x=34\)
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`
`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`
`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`
`B=1-(1-1/2023)`
`B=1-1+1/2023=1/2023`
Lạy ông đi quá lạy bà đi lại, ai có thể cho con xin cái đáp án k:>
a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)
b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)
c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)
d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
Chọn A