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a)đặt tên biểu thức là C . Ta có :
C = 1 + 4 + 42 + 43 + ... + 42012
C = ( 1 + 4 + 42 ) + ( 43 + 44 + 45 ) + ... + ( 42010 + 42011 + 42012 )
C = 21 + 43 . ( 1 + 4 + 42 ) + ... + 42010 . ( 1 + 4 + 42 )
C = 21 + 43 . 21 + ... + 42010 . 21
C = 21 . ( 1 + 43 + ... + 42010 )
=> C chia hết cho 21
b) đặt tên biểu thức là B . Ta có :
B = 1 + 7 + 72 + ... + 7101
B = ( 1 + 7 ) + ( 72 + 73 ) + ... + ( 7100 + 7101 )
B = 8 + 72 . ( 1 + 7 ) + ... + 7100. ( 1 + 7 )
B = 8 + 72 . 8 + ... + 7100 . 8
B = 8 . ( 1 + 72 + ... + 7100 )
=> B chia hết cho 8
tương tự
a) 13.(23+22)-3.(17+28)
= 13.45 - 3.45
= 45(13-3)
= 45.10 =450
b) -48+48.(-78)+48(-21)
= -48 + 48(-78 - 21)
= -48 + 48.(-99)
= -48 + ( -4752)
= -4800
c) 135.(171-123)-171.(135-123)
= 135.171 - 135.123 - 171.135 + 171.123
= (135.171 - 171.135) - (135.123 - 171.123)
= - 123(135 - 171) = (-123).(-36)= 4428
1) \(1+4+4^2+4^3+...+4^{2012}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2010}+4^{2011}+4^{2012}\right)\)
\(=21+21\cdot4^3+...+21\cdot4^{2010}\)
\(=21\cdot\left(1+4^3+...+4^{2010}\right)\) chia hết cho 21
2) \(1+7+7^2+7^3+...+7^{101}\)
\(=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{100}+7^{101}\right)\)
\(=8+8\cdot7^2+...8\cdot7^{100}\)
\(=8\cdot\left(1+7^2+...+7^{100}\right)\) chia hết cho 8
3) CM chia hết cho 5:
\(2+2^2+2^3+2^4+...+2^{100}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{98}+2^{100}\right)\)
\(=5\cdot2+5\cdot2^2+...+5\cdot2^{98}\)
\(=5\cdot\left(2+2^2+...+2^{98}\right)\) chia hết cho 5
CM chia hết cho 31:
\(2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\cdot31+...+2^{96}\cdot31\)
\(=31\cdot\left(2+...+2^{96}\right)\) chia hết cho 31
a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
a) 2(x-51) = 2^0.2^4 + 5.2^2
2(x-51) = 2^2(2^2+5)
2(x-51) = 36
x-51 = 18
x = 69
Vậy: x = 69
b) 225 : (x-19) = 25
x-19 = 9
x = 28
Vậy: x = 28
c) 4. (x-3)= 7^2-1^0
4. (x-3)= 48
x-3 = 12
x = 15
Vậy: x = 15
d) 3^3(x+4) - 3.5^2= 15.2^2
27 (x+4) - 75 = 60
27 (x+4) = 135
x+4 = 5
x = 1
Vậy: x = 1
e) 2x-7^2 = 5.3^2
2x - 49 = 45
2x = 94
x = 47
Vậy: x = 47
f) 2^3.5^2-(2x+6) = 4^3
200 - 2x - 6 = 64
194 - 2x = 64
2x = 130
x = 65
Vậy: x = 65
g) 135 - 5(x+4) = 35
5(x+4) = 100
x + 4 = 20
x = 16
Vậy: x = 16
h) 75 + 9(x-8) = 318
9(x-8) = 243
x-8 = 27
x = 35
Vậy: x = 35
\(A=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(=98.42-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(=98.42-\left[50\left(10:2+9\right)\right]\)
\(=98.42-\left(50.14\right)\)
\(=4116-700=3416\)
\(B=-80-\left[-130-\left(12-4\right)^2\right]+2008^0\)
\(=-80-\left(-130-8^2\right)+1\)
\(=-80-\left(-130-64\right)+1\)
\(=-80+130+64+1\)
\(=115\)
\(C=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(=1024:16+140:\left(38+32\right)-7^2\)
\(=64+140:70-49\)
\(=64+2-49=17\)
\(D=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0\)
\(=0\)
\(E=100+98+96+....+4+2-97-95-....-3-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(2-1\right)+\left(1-0\right)\)
\(=100+1+1+...+1+1\)
Vì lập được 49 cặp nên sẽ có 49 số 1
\(\Rightarrow E=100+1.49=100+49=149\)