A= \(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+....+\(\dfrac{1}{39800}\)
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GL
24 tháng 8 2023
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
GL
24 tháng 8 2023
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
AH
Akai Haruma
Giáo viên
10 tháng 3 2023
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
2
8 tháng 5 2022
`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`
`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`
`186+124-155 < 744x < -144+1632`
`155 < 744x < 1488`
`155:744 < 744x:744 < 1488:744`
`5/24 < x < 2`
Vậy `5/24 < x < 2`
__________________________________________________
`b)[-2]/5 < x/15 < 1/6`
`[-12]/30 < [2x]/30 < 5/30`
`-12 < 2x < 5`
`-12:2 < 2x:2 < 5:2`
`-6 < x < 5/2`
Vậy `-6 < x < 5/2`
8 tháng 5 2022
Giải:
a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)
x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)
x = 1
b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)
=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;2
TT
8 tháng 12 2021
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)
\(A=\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{39800}\)
\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)
\(A=\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{39800}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\dfrac{199}{200}\)