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\(A=2+2^2+2^3+...+2^{2016}\)
\(\Rightarrow2A=2\left(2+2^2+2^3+...+2^{2016}\right)\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{2017}\)
\(\Rightarrow2A-A=2^2+2^3+...+2^{2017}-2-2^2-2^3-...-2^{2016}\)
\(\Rightarrow A=2^{2017}-2\)
\(B=12+12^2+12^3+...+12^{100}\)
\(\Rightarrow12B=12\left(12+12^2+12^3+12^{100}\right)\)
\(\Rightarrow12B=12^2+12^3+...+12^{101}\)
\(\Rightarrow12B-B=12^2+12^3+...+12^{101}-12-12^2-12^3-...-12^{100}\)
\(\Rightarrow11B=12^{101}-12\)
\(\Rightarrow B=\dfrac{12^{101}-12}{11}\)
\(C=5+5^2+5^3+...+5^{32}\)
\(\Rightarrow5C=5\left(5+5^2+...+5^{32}\right)\)
\(\Rightarrow5C=5^2+5^3+...+5^{33}\)
\(\Rightarrow5C-C=5^2+5^3+...+5^{33}-5-5^2-...-5^{32}\)
\(\Rightarrow4C=5^{33}-5\)
\(\Rightarrow C=\dfrac{5^{33}-5}{4}\)
a) \(A=2+2^2+2^3+...+2^{2016}\)
\(2A=2^2+2^3+2^4+...+2^{2017}\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(\Rightarrow A=\left(2^2+2^3+2^4+...+2^{2016}\right)+2^{2017}-2-\left(2^2+2^3+2^4+...+2^{2016}\right)\)
\(\Rightarrow A=2^{2017}-2\)
Vậy \(A=2^{2017}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=129-128-127-126-125-124-123-122-121-120
A.12=1210-129-128-127-126-125-124-123-122-120
A.11=(1210-129-128-127-126-125-124-123-122-120)-(129-128-127-126-125-124-123-122-120)
A.11=1210-12
A=1210.11-11.12
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\(=2-\left\{8-\left[-7+3-7-12\right]+32\right\}\)
\(=2-\left\{40-\left[-23\right]\right\}=2-63=-61\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2-{3+5-[-12+5-(-3+7)-12]+32]
= 2-{3+5-[-12+5- 4 -12]+32]
=2-{3+5-[-7- 4 -12]+32]
= 2-{3+5-[-11 -12]+32]
= 2-{3+5-1+32]
= 2-{8-1+32]
=2-{7+32]
=2-32
=-30
2-{3+5-[-12+5-(-3+7)-12]+32 =2-{3+5-[-12+5+3-7-12]+32} =2-{3+5+12-5-3+7+12+32} =2-{(3-3)+(5-5)+(12+12+7+32)} =2-{0+63}=2-63=-61
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\dfrac{5}{12}+\dfrac{5}{3}+\dfrac{2}{12}+\dfrac{2}{3}\)
\(=\left(\dfrac{5}{12}+\dfrac{2}{12}\right)+\left(\dfrac{5}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{12}+\dfrac{7}{3}\)
\(=\dfrac{7}{12}+\dfrac{28}{12}\)
\(=\dfrac{35}{12}\)
\(b)\dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}-\dfrac{5}{9}.\dfrac{3}{13}\)
\(=\dfrac{5}{9}.1.\dfrac{7}{13}+\dfrac{5}{9}-\dfrac{5}{9}.\dfrac{3}{13}\)
\(=\dfrac{5}{9}.\left(1+\dfrac{7}{13}-\dfrac{3}{13}\right)\)
\(=\dfrac{5}{9}.\dfrac{17}{13}\)
\(=\dfrac{85}{117}\)
\(12^5\div12^3-12^2=12^2-12^2=0\)
125:123-122
=125-3-122
=122-122
=0