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\(A=\frac{11^{40}+1}{11^{43}+1}\)
\(11^3A=\frac{11^3\left(11^{40}+1\right)}{11^{43}+1}=\frac{11^{43}+1331}{11^{43}+1}=\frac{11^{43}+1+1330}{11^{43}+1}=\frac{11^{43}+1}{11^{43}+1}+\frac{1330}{11^{43}+1}=1+\frac{1330}{11^{43}+1}\)
\(B=\frac{11^{41}+1}{11^{44}+3}\)
\(11^3B=\frac{11^3\left(11^{41}+1\right)}{11^{44}+3}=\frac{11^{44}+1331}{11^{44}+3}=\frac{11^{44}+3+1328}{11^{44}+3}=\frac{11^{44}+3}{11^{44}+3}+\frac{1328}{11^{44}+3}=1+\frac{1328}{11^{44}+3}\)
Ta có: \(\frac{1330}{11^{43}+1}>\frac{1330}{11^{44}+3}>\frac{1328}{11^{44}+3}\)
\(\Rightarrow\frac{1330}{11^{43}+1}>\frac{1328}{11^{44}+3}\)
\(\Rightarrow1+\frac{1330}{11^{43}+1}>1+\frac{1328}{11^{44}+3}\)
\(\Rightarrow11^3A>11^3B\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
P/s: Hoq chắc :<
Ta có: A=47.31 +67.41 +910.41 +710.57
=2035.31 +3035.41 +4550.41 +3550.57
=5(431.35 +635.41 +941.50 +750.57 )
=5(131 −135 +135 −141 +141 −150 +150 −157 )
=5(131 −157 )
B=719.31 +519.43 +323.43 +1123.57
=1438.31 +1038.43 +646.43 +2246.57
=2(731.38 +538.43 +343.46 +1146.57 )
=2(131 −138 +138 −143 +143 −146 +146 −157 )
=2(131 −157 )
⇒AB =5(131 −157 )2(131 −157 ) =52
B = (4^1 + 4^2) + (4^3 +4^4) + ... + (4^299 + 4^300)
= 4(1+4)+4^3(1+4)+...+4^299(1+4)
= 4.5+4^3 .5 +...+4^299. 5
= 5.(4+4^3+...+4^299) chia hết cho 5
\(B=4^1+4^2+4^3+4^3+...+4^{300}\\=(4+4^2)+(4^3+4^4)+(4^5+4^6)+...+(4^{299}+4^{300})\\=4\cdot(1+4)+4^3\cdot(1+4)+4^5\cdot(1+4)+...+4^{299}\cdot(1+4)\\=4\cdot5+4^3\cdot5+4^5\cdot5+...+4^{299}\cdot5\\=5\cdot(4+4^3+4^5+...+4^{299})\)
Vì \(5\cdot(4+4^3+4^5+...+4^{299}) \vdots 5\)
nên \(B \vdots 5\)
Ta có: \(4M=4^2+4^3+4^4+...+4^{2021}+4^{2022}\)
\(\Rightarrow4M-M=4^2+4^3+4^4+...+4^{2021}+4^{2022}-\left(4+4^2+4^3+...+4^{2020}+4^{2021}\right)\)
\(\Leftrightarrow3M=4^{2022}-4\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
\(4M=4^2+4^3+...+4^{2022}\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
A = 40 + 41 + 42+ 43 + ....+42020
4A = 41 + 42 + 43 +....+42020 + 42021
4A - A = 42021 -1
3A = 42021 - 1
A = (42021 - 1):3
B - A = 42021 : 3 - (42021 -1) : 3
B - A = ( 42021 - 42021 + 1): 3
B - A = 1/3