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Bài 1:
$A=2^1+2^2+2^3+2^4$
$2A=2^2+2^3+2^4+2^5$
$\Rightarrow 2A-A=2^5-2^1$
$\Rightarrow A=2^5-1=32-1=31$
----------------------------
$B=3^1+3^2+3^3+3^4$
$3B=3^2+3^3+3^4+3^5$
$\Rightarrow 3B-B = 3^5-3$
$\Rightarrow 2B = 3^5-3\Rightarrow B = \frac{3^5-3}{2}$
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$C=5^1+5^2+5^3+5^4$
$5C=5^2+5^3+5^4+5^5$
$\Rightarrow 5C-C=5^5-5$
$\Rightarrow C=\frac{5^5-5}{4}$
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a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\frac{1}{10}< =x< =\frac{3}{5}\)
\(\frac{-4}{9}< x< =\frac{2}{3}\)
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A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
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a) \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{34}\right)\)
\(A=\frac{2}{3}\cdot\frac{33}{34}=\frac{11}{17}\)
b) \(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)
\(B=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\) ( 3/1 = 6/2; 6/6=3/3;..)
\(B=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(B=6.\left(1-\frac{1}{21}\right)=6\cdot\frac{20}{21}=\frac{40}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`3x-16:2^3=31`
`=>3x-16:8=31`
`=>3x-2=31`
`=>3x=31+2`
`=>3x=33`
`=>x=11`
__
`2^10:2^8+3[4.7+3.4]`
`=2^2+3[4(3+7)]`
`=4+3[4.10]`
`=4+3.40`
`=4+120`
`=124`
__
`4^6:4^3-2^2 . 2^3`
`=4^3-2^5`
`=64-32`
`=32`
__
`141+2^5 . 2^4-3^1 . 3^2`
`=141+2^9-3^3`
`=141+512-9`
`=644`
__
`x+2^5:2^4=4.4^2`
`=>x+2=4^3`
`=>x=64-2`
`=>x=62`
\(A=1+3+3^2+...+3^{31}\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{29}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{29}.13\)
\(=13\left(1+3^3+...+3^{29}\right)⋮13\)
\(\Rightarrow A⋮13\)
\(A=1+3+3^2+3^3+...+3^{31}\)
\(\Rightarrow A=3^0+3^1+3^2+...+3^{31}\)
\(\text{Số số hạng của A là : ( 31 - 0 ) + 1 = 32 ( số )}\)
\(\text{Chia A làm 10 nhóm và dư 2 số }\)
\(\Rightarrow A=\left(3^0+3^1+3^2\right)+...+\left(3^{27}+3^{28}+3^{29}\right)+3^{30}+3^{31}\)
\(\Rightarrow A=13+...+3^{27}.13+3^{30}\left(1+3\right)\)
\(\Rightarrow A=13\left(1+...+3^{27}\right)+3^{30}.4\)
\(\Rightarrow A⋮13\)