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a: =25x100-150=2500-150=2350
c: \(=520:\left\{515\cdot25\right\}\)
=104/2575
9*8*48+7*4*48+72*52+2*52*14
=72*48+28*48+72*52+28*52
=72 * ( 48 + 52 ) + 28 * ( 48 + 52 )
=72 * 100 + 28 * 100
=100 * ( 72 + 28 )
= 100 * 100
=10000
9 x 8 x 48 + 7 x 4 x 48 + 72 x 52 + 2 x 52 x 14
= 72 x 48 + 28 x 48 + 72 x 52 + 2 x 14 x 52
= ( 72 + 28 ) x 48 + 72 x 52 + 28 x 52
= ( 72 + 28 ) x 48 + ( 72 + 28 ) x 52
= 100 x 48 + 100 x 52
= 100 x ( 48 + 52 )
= 100 x 100
= 10000
a) - 72 . ( 15 - 49 ) + 15 . ( - 56 + 72 ) = -72 . -34 +15 . 16
= 2448 + 240
= 2688
b) 1532 + ( -168 ) + ( - 1432 ) + ( - 14 ) + 123 = 1364 + ( -1432) + ( -14) +123
= -68 + ( -14 ) +123
= -82 + 123 = 41
c) 75 - 5 . ( 15 - 40 ) - ( - 60 ) = 75 - 5 . (-25 )-(-60)
= 75 - (-125) - ( -60 )
= 200 - ( -60 ) = 260
d) | 31 -17 | - | 15 - 52 | = | 14 | - | -37 |
= 14 - 37
= -23
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)