Cho :\(A=1+6+6^2+...+6^{59}\)
CMR:
\(a,A⋮43\)
\(b,A⋮7\)
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NT
0
DT
1
DT
5 tháng 7 2018
a) \(8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}=2^{17}\left(2^4-2\right)=2^{17}.\left(16-2\right)=2^{17}.14⋮14\)
b) \(10^6-5^7=5^6.2^6-5^7=5^6.\left(2^6-5\right)=5^6.\left(64-5\right)=5^6.59⋮59\)
20 tháng 9 2017
a) 41. 36+ 59. 90 + 41. 84 + 59. 30
= 41 . (36 + 84) + 59 . (90 + 30)
= 41 . 120 + 59 . 120
= 120 . ( 41 + 59 )
= 120 . 100
= 12000
b) 4. 51 . 7 + 2 . 86 .7 + 6 . 4 . 7
= 28 . 51 + 2 . 43 . 2 . 7 + 6 . 4 . 7
= 28 . 51 + 28 . 43 + 28 . 6
= 28 . ( 51 + 43 + 6 )
= 28 . 100
= 2800
c) ( 4.7 ) . 5 + ( 4.7 ) . 43 + ( 4.7) - 6
= 28 . 5 + 28 .43 + 28 - 6
= 28 . (5 + 43 + 1) - 6
= 28 . 49 - 6
= 1372 - 6
= 1366.
d) 53 . 39 + 47 . 39 - 53 . 21 - 47 . 21
= 53 . (39 - 21) + 47. (39 - 21)
= (39 - 21) . ( 53 + 47 )
= 18 . 100
= 1800.
13 tháng 8 2016
a,24.57+43.24
=24.(57+43)
=24.100
=2400
b,12.19+12
=12.(19+1)
=12.20
=12.10.2
=120.2
=240
c,43.27+93.43+57.61+59.57
=43.(27+93)+57.(61+59)
=43.120+57.120
=120.(43+57)
=120.100
=12000
d,64.6+81.4+17.6
=64.6+27.3.2.2+17.6
=64.6+54.6+17.6
=(64+54+17).6
=135.6
=810
13 tháng 8 2016
a) 24 . 57 + 43 . 24
= 24 . (57 + 43)
= 24 . 100
= 2400
b) 12 . 19 + 12
= 12 . (19 + 1)
= 12 . 20
=240
c) 43 . 27 + 93 . 43 + 57 . 61 +59 . 57
= 43 . ( 27 + 93) + 57 . (61 + 59)
= 43 . 120 + 57 . 120
= 120 . (43 + 57)
= 120 . 100
= 12000
LD
1
23 tháng 1 2021
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
23 tháng 1 2021
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
A = ( 1 + 6 + 6^2 ) + ( 6^3 + 6^4 + 6^5 ) + ... + ( 6^57 + 6^58 + 6^59 )
= 1( 1 + 6 + 6^2 ) + 6^3( 1 + 6 + 6^2 ) + ... + 6^57( 1 + 6 + 6^2 )
= 1.43 + 6^3.43 + ... + 6^57.43
= 43( 1 + 6^3 + ... + 6^57 )
=> A chia hết cho 43
A = ( 1 + 6 ) + ( 6^2 + 6^3 ) + ... + ( 6^58 + 6^59 )
= 1( 1 + 6 ) + 6^2( 1 + 6 ) + ... + 6^58( 1 + 6 )
= 1.7 + 6^2.7 + ... + 6^58.7
= 7( 1 + 6^2 + ... + 6^58 )
=> A chia hết cho 7