1+-1x3+2-1:7+4-6+9=?
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8 tháng 4 2021
Giải:
A=5/9+2/15-6/9
=(5/9-6/9)+2/15
= -1/9 + 2/15
= 1/45
B=2/7-3/8+4/7+1/7-5/8+5/15
= (2/7+4/7+1/7) + (-3/8-5/8) +1/3
= 1+ (-1) +1/3
=1/3
C=3/5+1/15+1/57+1/3-2/9-3/4-1/36
=9/15+1/15+1/57+19/57-8/36-27/36-1/36
=(9/15+1/15)+(1/57+19/57)+(-8/36-27/36-1/36)
=2/3+20/57+(-1)
=58/57+(-1)
=1/57
D=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1/1-1/100
=99/100
Câu E mình ko biết làm nhé!
XO
19 tháng 8 2020
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
PT
3
C
29 tháng 5 2019
\(\left(x\cdot2,4-4,2\right)\div x=1\)
\(\Rightarrow x\cdot2,4-4,2=x\)
\(\Rightarrow x\cdot2,4=x+4,2\)
\(\Rightarrow\frac{12x}{5}=\frac{5x+21}{5}\)
\(\Rightarrow12x=5x+21\)
\(\Rightarrow12x-5x=21\)
\(\Rightarrow7x=21\Rightarrow x=\frac{21}{7}=3\)
Vậy x = 3
C
29 tháng 5 2019
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{x+1}{2x+2}\)
O
3 tháng 5 2022
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
O
3 tháng 5 2022
Sửa câu b)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
Đặt \(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(2A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(2A=1-\dfrac{1}{9}\)
\(2A=\dfrac{9}{9}-\dfrac{1}{9}\)
\(2A=\dfrac{8}{9}\)
\(A=\dfrac{8}{9}:2\)
\(A=\dfrac{8}{18}\)
\(A=\dfrac{4}{9}\)
Vậy : \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}=\dfrac{4}{9}\)
16 tháng 8 2019
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
16 tháng 8 2019
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
CM
14 tháng 6 2018
5 + 4 = 9 1 + 6 = 7 4 + 2 = 6 9 + 1 = 10 2 + 7 = 9
9 - 5 = 4 7 - 1 = 6 6 - 4 = 2 10 - 9 = 1 9 - 2 = 7
9 - 4 = 5 7 - 6 = 1 6 - 2 = 4 10 - 1 = 9 9 - 7 = 2
9 tháng 8 2018
a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
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