So sánh:
1. 102 + 112 + 122 và 132 + 142
2. (30 + 25)2 và 3025
3. 37.(3 + 7) và 33 + 73
4. 48.(4 + 8) và 43 + 83
5. A = 20 + 21 + 22 + 23 + … + 22010 và B = 22011 - 1.
6. B = 2009.2011 và B = 20102.
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\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)
\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)
\(\Rightarrow A=2^{2011}-1=B\)
\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)
\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)
\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
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:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50=1275
a) \(\sqrt{3}+5=\sqrt{3}+\sqrt{25}>\sqrt{2}+\sqrt{11}\)
b) \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
c) \(4+\sqrt{33}=\sqrt{16}+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
d) \(\sqrt{48}+\sqrt{120}< \sqrt{49}+\sqrt{121}=7+11=18\)
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
Cảm ơn Trần Quỳnh Mai nhé!