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1. A - B = 40+ 3/8 + 7/82 + 5/83 + 32/85 - (24/82 + 40+ 5/82 + 40/84 + 5/84 )
= 40.85/85 + 3.84/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 40.85/85 + 24.83/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 + 32/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 -8/85 - 5.83/85 - 40.8/85
= 2.83/85 + 5.82/85 - 40.8/85 - 8/85
= 2.83/85 + 40.8/85 - 40.8/85 - 8/85
= 2.83/85 - 8/85 > 0
Vay A > B
Bài 1:
a)12,5 x (-5/7) + 1,5 x (-5/7)
=-5/7*(12,5+1,5)
=-5/7*14
=-10
b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)
=-1/4*(68/11+42/11)
=-1/4*10
=-5/2
c tương tự
d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)
Bài 2:
a)Ta có:
2800=(28)100=256100
8200=(82)100=64100
Vì 256100>64100 =>2800>8200
b)Ta có:
1245=(123)15=172815
Vì 62515<172815 =>62515<1245
\(11^{12}< 11^{13}\)
\(7^4< 8^4\)
\(3^4>4^3\)
\(2^6>6^2\)
\(5^{15}>2^{30}\)
a,12,5x(-5/7)+1,5x(-5/7)
=-125/14+-15/14
=-10
2,2mu800>8 mu 200
6254 lon hon 12
\(11^{12}< 11^{13}\)
\(7^4< 8^4\)
\(\left(6-5\right)^{132}=1^{132}=1\)
\(\left(7-6\right)^{543}=1^{543}=1\)
\(\Rightarrow\left(6-5\right)^{132}=\left(7-6\right)^{543}\)
\(37\left(3+7\right)=37.10=370\)
\(3^3+7^3=27+343=370\)
\(\Rightarrow37\left(3+7\right)=3^3+7^3\)
\(147\left(14+7\right)=147.21=3087\)
\(14^3+7^3=2744+343=3087\)
\(\Rightarrow147\left(14+7\right)=14^3+7^3\)
Bài 1:
a) Ta có: 536=(53)12=12512
1124=(112)12=12112
Vì 12512>12112
=>536>1124
b) Ta có: 6255=(54)5=520
1257=(53)7=521
Vì 520<521
=>6255<1257
c) Ta có: 32n=(32)n=9n
23n=(23)n=8n
Vì 9n>8n
=>32n>23n
d) Ta có: 6.522=(1+5).522=523+522>523
e) S=1+2+22+23+...+22005
2S=2+22+23+24+...+22006
=>2S-S=(2+22+23+24+...+22006) - (1+2+22+23+...+22005)
=>S=22006-1<22014<5.22014
Cậu cho tớ 3 tớ sẽ làm 2 bài còn lại cho cậu
1, Ta có : \(10^2+11^2+12^2=100+121+144=365\)
\(13^2+14^2=169+196=365\)
Vì : \(365=365\Rightarrow10^2+11^2+12^2=13^2+14^2\)
Vậy \(10^2+11^2+12^2=13^2+14^2\)
2, \(\left(30+25\right)^2=30^2+25^2=900+625=1525\)
Vì : \(1525< 3025\Rightarrow\left(30+25\right)^2< 3025\)
Vậy \(\left(30+25\right)^2< 3025\)
3, \(37\left(3+7\right)=37.10=370\)
\(3^3+7^3=\left(3+7\right)^3=10^3=1000\)
Vì : \(370< 1000\Rightarrow37\left(3+7\right)< 3^3+7^3\)
Vậy \(37\left(3+7\right)< 3^3+7^3\)
4, \(48\left(4+8\right)=48.12=576\)
\(4^3+8^3=\left(4+8\right)^3=12^3=1728\)
Vì : \(576< 1728\Rightarrow48\left(4+8\right)< 4^3+8^3\)
Vậy \(48\left(4+8\right)< 4^3+8^3\)
5, \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2011}\right)-\left(1+2+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-1\)
Vì : \(2^{2011}-1=2^{2011}-1\Rightarrow A=B\)
Vậy A = B
6, Ta có : \(A=2009.2011=2009.\left(2010+1\right)\)
\(=2009.2010+2009\)
\(B=2010^2=2010.2010\)
\(=2010.\left(2009+1\right)=2010.2009+2010\)
Vì : \(2010.2009+2009< 2010.2009+2010\Rightarrow A< B\)
Vậy A < B
a) <
c) <
a] 2^6 va 6^2
2^6>6^2
b] 11^12 va 11^14
11^12<11^14
c] 7^4 va 8^4
7^4< 8^4
d] [ 6-5]^217 va [8-7] ^123
[5-5]^217 = [8-7]^123
tk nha