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Bài 2:
\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3
Vậy: A ⋮ 3
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\(A=2+2^2+...+2^{60}\)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)
\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)
\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7
Vậy: A ⋮ 7
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\(A=2+2^2+...+2^{60}\)
\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)
\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)
\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)
\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5
Vậy: A ⋮ 5
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Bài 48:
1/ 17(-84) + 17(-16) = 17 \(\left[\left(-84\right)+\left(-16\right)\right]\) = 17. (-100) = -1700
2/ 15.58 - 15.48 = 15.(58 - 48) = 15 . 10 = 150
3/ -37.86 + 37.76 = 37 .(-86 + 76) = 37. (-10) = -370
4/ 1975 .(-115) + 1975.15 = 1975. (-115 + 15) = 1975 . (-100) = -197500
5/ 79.89 - 79.(-11) = 79. (89 + 11) = 79 . 100 = 7900
6/ 48.195 - 48 . 95 = 48. (195 - 95) = 48 . 100 = 4800
7/ 157.17 - 157.7 = 157. (17 - 7) = 157 . 10 = 1570
8/ 15(-176) + 15.76 = 15. (-176 + 6) = 15 . 100 = 1500
9/ 47.(-147) - 47 .(-47)= 47 (-147 + 47) = 47. (-100) = -4700
10/ 153.177 - 153.77 = 153. (177 - 77) = 153. 100 = 15300
11/ 15. (4 - 7) - 15. (5 - 3)
= 15. (-3) - 15. 2
= 15. (-3 - 2)
= 15. (-5)
= -75
12/ 73. (8 - 59) - 59. (8 - 73)
= 73. 8 - 73. 59 - 59. 8 + 73. 59
= 8. (73 - 59) + 73. (59 - 59)
= 8. (73 - 59) + 0
= 8. 14
= 112
13/ 159. (18 - 59) - 59. (18 - 159)
= 159. 18 - 159. 59 - 59. 18 + 59. 159
= 18.(159 - 59) - 59. (159 - 159)
= 18.(159 - 59) - 0
= 18. 100
= 1800
14/ -145. (13 - 57) + 57. (10 - 145)
= -145. 13 + 145. 57 + 57. 10 - 57. 145
= -145. 13 + 57. 10
= -1885 + 570
= 2455
Những câu còn lại tương tự
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Bài 2:
Cả mảnh vải dài:
\(25:\dfrac{3}{5}=\dfrac{125}{3}\left(m\right)\)
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a: =>x+2/3=4/60=1/15
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
b: =>x/3=14/21-3/21=11/21
=>x=11/7
\(x+\dfrac{2}{3}=-\dfrac{1}{12}.\dfrac{-4}{5}.\\ \Leftrightarrow x+\dfrac{2}{3}=\dfrac{1}{15}.\\ \Leftrightarrow x=\dfrac{-3}{5}.\\ \dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\\Leftrightarrow \dfrac{7x}{21}=\dfrac{11}{21}.\\\Leftrightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}. \)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\dfrac{-5}{9}+\dfrac{4}{9}=\dfrac{-1}{9}\)
b: \(=\dfrac{5}{17}+\dfrac{-1}{17}\cdot3=\dfrac{5}{17}-\dfrac{3}{17}=\dfrac{2}{17}\)
c: \(=\dfrac{1}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{1}{5}=\dfrac{1}{5}-\dfrac{1}{5}=0\)
d: =5,17-2,24-5,17+3,24=1
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a: Ta có: \(\dfrac{n\left(n+1\right)}{2}=231\)
\(\Leftrightarrow n^2+n-462=0\)
\(\Leftrightarrow n^2+22n-21n-462=0\)
\(\Leftrightarrow\left(n+22\right)\left(n-21\right)=0\)
\(\Leftrightarrow n=21\)
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Bài 11:
a: x/5=2/10
=>x/5=1/5
hay x=1
b: x/-3=-5/15
=>x/3=1/3
hay x=1