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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6 2024
a; [6.(- \(\dfrac{1}{3}\))3 - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)
= [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] - ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\))
= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)
= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)
= - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)
= \(\dfrac{-8}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6 2024
b; (63 + 3.62 + 33): 13
= (216 + 3.36 + 27) : 13
= (216 + 108 + 27): 13
= (324 + 27): 13
= 351 : 13
= 27
15 tháng 7 2017
a) Ta thấy :
U1 = 1 . 3 ; U2 = 2 . 4 ; U3 = 3 . 5 ; ... ; Un = n . ( n + 2 )
c) U1 = 1 ; U2 = 1 + 2 ; U3 = 1 + 2 + 3 ; U4 = 1 + 2 + 3 + 4 ; U5 = 1 + 2 + 3 + 4 + 5 ; ... : Un = 1 + 2 + 3 + ... + n
d) 2 + 3 = 5 ; 5 + 5 = 10 ; 10 + 7 = 17 ; 17 + 9 = 26 ; ...
f) 4 = 1 . 4 ; 28 = 4 . 7 ; 70 = 7 . 10 ; 130 = 10 . 13 ; 208 = 13 . 16 ; ...
g) 2 + 3 = 5 ; 5 + 4 = 9 ; 9 + 5 = 14 ; 14 + 6 = 20 ; ...
i) 2 + 6 = 8 ; 8 + 12 = 20 ; 20 + 20 = 40 ; 40 + 30 = 70 ; ...
TM
1
XO
3 tháng 3 2020
C = \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{210}\right)=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{209}{210}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}=\frac{2.2\left(2.3.4...19\right)\left(5.6...22\right)}{\left(2.3.4..20\right)\left(3.4.5..21\right)}=\frac{4.22}{19.3.4}=\frac{22}{57}\)
TT
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 6 2024
a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)
\(x\) = \(\dfrac{29}{30}\)
Vậy \(x\) = \(\dfrac{29}{30}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 6 2024
b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???
c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\)
\(x\) - \(\dfrac{1}{2}\) = 2
\(x\) = 2 + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
S
1 tháng 7 2018
\(\frac{1}{2}B=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)
\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\Rightarrow B=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)
1 tháng 7 2018
\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(B=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(B=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(B=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(B=2.\frac{3}{16}=\frac{6}{16}=\frac{3}{8}\)
\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...........+\dfrac{1}{120}\)
\(\Leftrightarrow C=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+..........+\dfrac{2}{240}\)
\(\Leftrightarrow C=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...........+\dfrac{1}{240}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+............+\dfrac{1}{15.16}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow C=2.\dfrac{7}{16}=\dfrac{7}{8}\)
\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{120}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{120}\right)\)
\(\dfrac{1}{2}C=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{240}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{15}-\dfrac{1}{16}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}-\dfrac{1}{16}\)
\(\dfrac{1}{2}C=\dfrac{7}{16}\)
\(C=\dfrac{7}{8}\)