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Ta có :
\(B=4+2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow\) \(B-4=2^2+2^3+2^4+...+2^{2016}\)
\(\Rightarrow\) \(2\left(B-4\right)=2^3+2^4+2^5+...+2^{2017}\)
\(\Rightarrow\) \(2\left(B-4\right)-\left(B-4\right)=B-4=2^{2017}-2^2\)
\(\Rightarrow\) \(B=2^{2017}-2^2+4=2^{2017}\)
\(\Rightarrow\) \(A=B=2^{2017}\)
Vậy \(A=B\)
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`[x-3]/5=[2x-6]/10`
`[2(x-3)]/10=[2x-6]/10`
`2x-6=2x-6`
`2x-2x=-6+6`
`0x=0` (LĐ)
Vậy `x in RR`
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(x+1)+(2x+3)+(3x+5)+...+(100x+199)=30200
=(x+2x+3x+..........+100x)+(1+3+5+......+199)=4950
=x.(1+2+3+.........+100)+(1+3+5+.......+199)=4950
=>5050x+10000=30200
=>5050x=30200-4950
=>5050x=25250
=>x=25250:5050
=>x=5
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a : vì \(\dfrac{354}{758}< 1;\dfrac{123}{56}>1\\ \Rightarrow\dfrac{354}{758}< \dfrac{123}{56}\)
b:\(\dfrac{-18}{2021}< 0;\dfrac{75}{2022}>0\\ \Rightarrow\dfrac{-18}{2021}< \dfrac{75}{2022}\)
a) Do 354/758<1 ;123/56>1
=>354/758<123/56
b) -18/2021<0;72/2022>0
=>-18/2021<72/2022
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\(\dfrac{5}{-6}=\dfrac{-55}{66};\dfrac{-10}{11}=\dfrac{-60}{66}\Rightarrow\dfrac{-55}{66}>\dfrac{-60}{66}\Rightarrow\dfrac{5}{-6}>\dfrac{-10}{11}\\ \dfrac{-3}{20}=\dfrac{-45}{300};\dfrac{2}{-15}=\dfrac{-40}{300}\Rightarrow\dfrac{-45}{300}< \dfrac{-40}{300}\Rightarrow\dfrac{-3}{20}< \dfrac{2}{-15}\\ -0,305>-0,36\)
\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)
Ta có \(:\)
\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)
\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)
Do \(55 < 60\)
\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)
Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
Cộng vế với vế ta được
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)
mà 39/20 < 8/7 => T < 8/7
chắc là bé hơn