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\(Q=\frac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Ta có :
\(\hept{\begin{cases}\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\\\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\\\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\end{cases}}\)
\(\Rightarrow P>Q\)
Bài 1:
1. 36.7 + 34.37 + 19.100
= 34.(32.7 + 37) + 19.100
= 81.100 + 19.100
= 100.(81 + 19)
= 100.100
= 10000
2) 2.14.98+7.4.32-28.30
= 28.98 + 28.32 - 28.30
= 28. (98 + 32 - 30)
= 28.100
= 2800
3) (56.35+56.18):53
= [56.(35 + 18)] : 53
= 56.53:53
= 56
4) (158.129-158.39):28
= [158. (129 - 39)] : 28
= 158.90:28
= 5,675
1) 3.52 - 16 : 22
= 3 . 25 - 24 : 22
= 75 - 22
= 75 - 4
= 71
2) 23 .17 -23 . 14
= 23 . ( 17 - 14 )
= 8 . 3
= 24
3) 17.85 + 15.17 - 120
= 17 . ( 85 + 17 ) - 120
= 75 . 102 - 120
= 7650 - 120
= 7530
4) 20 - [30-(5 - 1)2]
= 20 - ( 30 - 42 )
= 20 - ( 30 - 16 )
= 20 - 14
= 6
5) 36 . 32 + 23 .22
= 38 + 25
= 6561 + 32
= 6593
6) 17 .85 + 15.17 - 120 + 20120
= 17 . ( 85 + 15 ) - 120 + 1
= 17 . 100 - 120 + 1
= 1700 - 120 + 1
= 1580 + 1
= 1581
7) 37. 24 + 37. 76 + 63 . 79 + 21. 63
= 37 . ( 24 + 76 ) + 63 . ( 79 + 21 )
= 37 . 100 + 63 . 100
= 100 . ( 37 + 63 )
= 100 . 100
= 10000
1) \(3.5^2-16.2^2=3.25-16.4=75-64=11\)
2) \(2^3.17-2^3.14=2^3.\left(17-14\right)=8.3=24\)
3) \(17.85+15.17-120=17.\left(85+15\right)-120\)
\(=17.100-120=1700-120=1580\)
4)\(20-\left[30-\left(5-1\right)^2\right]=20-\left[30-4^2\right]\)
\(=20-\left[30-16\right]=20-14=6\)
5) \(3^6.3^2+2^3.2^2=3^8+2^5=6561+32=6593\)
6) \(17.85+15.17-120+2012^0=17.\left(85+15\right)-120+1\)
\(=17.100-119=1700-119=1581\)
7) \(37.24+37.76+63.79+21.63\) \(=37.\left(24+76\right)+63.\left(79+21\right)\)
\(=37.100+63.100=100.\left(37+63\right)=100.100=10000\)
phần b tương tự phần a nên em làm câu a và c thôi :
a, \(M=1-2+2^2-2^3+...+2^{2012}\)
\(2M=2-2^2+2^3-2^4+...+2^{2013}\)
\(3M=2^{2013}+1\)
\(M=\frac{2^{2013}+1}{3}\)
c, \(E=2^{100}-2^{99}-2^{98}-...-1\)
\(E=2^{100}-\left(2^{99}+2^{98}+...+1\right)\)
đặt \(A=2^{99}+2^{98}+...+1\)
\(2A=2^{100}+2^{98}+...+2\)
\(2A-A=2^{100}-1\) hay \(A=2^{100}-1\)
ta có :
\(E=2^{100}-\left(2^{100}-1\right)\)
\(E=2^{100}-2^{100}+1=1\)
Vì: \(2^4\)có tận cùng là đặc biệt
Ta có: \(2^{2013}=2^{4.503+1}=\left(2^4\right)^{503}.2=\overline{....6}^{503}.2=\overline{....2}\)
\(\dfrac{5.4^2+16}{2^3}=\dfrac{16\left(5+1\right)}{2^3}=2.6=12\)
\(\dfrac{5^{16}}{5^{14}}+2^2.2^3=5^2+2^5=25+32=57\)
\(\dfrac{7^{2012}}{7^{2010}}-6^2=7^2-6^2=49-36=13\)
\(2^2.3+\dfrac{250}{5^2}=12+10=22\)
\(2.9.50-2012^0=9.100-1=899\)
\(\dfrac{123}{3}-\dfrac{4^3}{2^4}=41-\dfrac{4^2.4}{2^4}41-4=37\)