Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(\frac{1\div2003+1\div2004-1\div2005}{5\div2003+5\div2004-5\div2005}\) - \(\frac{2\div2002+2\div2003-2\div2004}{3\div2002+3\div2003-3\div2004}\)
Đơn giản đi hết ta sẽ còn:
\(\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
2.
Ta có:
Số khoảng cách của các số trong dãy là 23 = 8
=> Tổng của dãy dưới sẽ gấp 8 lần tổng dãy trên.
=> 3025 . 8 = 24200
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,\(A=2^{200}-2^{199}-2^{198}-...-2-1\)
\(2A=2^{201}-2^{200}-...-2^2-2\)
\(2A-A=A=2^{101}+1\)
b,\(b=2^3+4^3+...+20^3\)
\(b=2^3\left(1^3+2^3+...+10^3\right)\)
\(b=8.2025\)
\(b=16200\)
S = 2^2 + 4^2 +.. +20^2
= 2^2 . 1 +2^2 .2^2 +...+2^2 .10^2
= 2^2 ( 1+ 2^2 +...+10^2)
= 4 . S
= 4. 385
= 1540
a)410x220x830 b)56x53+32x33
=(22)10x220x(23)30 =56+3+32+3
=220x220x290 =59+35
=220+20+90 =1953125+243
=2130 =1953368.
c)13+23+33+43+53 d)225:324
=1+8+27+64+125 =225 :(25)4
=9+27+64+125 =225:220
=36+64+125 =225-20
=100+125 =25
=225
a)\(\left(2^2\right)^{10}.2^{20}.\left(2^3\right)^{30}=2^{20}.2^{20}.2^{90}=2^{130}\)
b)\(5^{6+3}+3^{2+3}=5^9+3^5\)
c)\(1^3+2^3+3^3+4^3+5^3=\left(1+2+3+4+5\right)^2=15^2=225\)
d)\(2^{25}:32^4=2^{25}:\left(2^5\right)^4=2^{25}:2^{20}=2^{25-20}=2^5=32\)
4
a)\(S=1+2+2^2+...+2^{10}\)
\(2S=2+2^2+2^3+...+2^{11}\)
\(2S-S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(S=2^{11}-1\)
b)\(S=1+3+3^2+...+3^6\)
\(3S=3+3^2+3^3+...+3^7\)
\(3S-S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)
\(2S=3^7-1\)
\(S=\frac{3^7-1}{2}\)
a.\(S=1+2+2^2+...+2^{10}\)
\(2S=2+2^2+2^3+...+2^{11}\)
\(\Rightarrow2S-S=S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(=2^{11}-1\)
b) \(S=1+3+3^2+...+3^6\)
\(3S=3+3^2+3^3+...+3^7\)
\(\Rightarrow3S-S=2S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)
\(2S=3^7-1\Rightarrow S=\frac{3^7-1}{2}\)
Ta có : S = 22 + 42 + 62 + .... + 202
= 12 . 22 + 22. 22 + 22 . 32 + .... + 22. 102
= 22.(12 + 22 + 32+ .... + 102)
Mà 12 + 22 + 32+ .... + 102 = 385
=> 22 . (12 + 22 + 32+ .... + 102) = 22 . 385 = 4 . 385 = 1540
Vậy S = 1540
S = 22 + 42 + 62 + ........ + 202
= 12. 22 + 22.22 + 32.22+ ........ + 102.22
= 22.(12 + 22 + 32+..........+ 102)
= 4.385
= 1540
\(S=2^3+4^3+6^3+20^3\)
\(=2^3\cdot1^3+2^3\cdot2^3+2^3\cdot3^3+.....+2^3\cdot10^3\)
\(=2^3\left(1^3+2^3+3^3+....+10^3\right)\)
\(=8\cdot\left(1^3+2^3+3^3+...+10^3\right)\)
Mà \(1^3+2^3+3^3+....+10^3=3025\Rightarrow S=8\cdot3025=24200\)
Vậy S=24200