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\(A=11,44+12,56-30,05+1024\)
\(=24-30,56+1024\)
\(=-6,05+1024\)
\(=1017,95\)
1,
\(\begin{array}{l}2,5.\left( {4,1 - 3 - 2,5 + 2.7,2} \right) + 4,2:2\\ = 2,5.\left( {4,1 - 3 - 2,5 + 14,4} \right) + 4,2:2\\ = 2,5.\left( {1,1 - 2,5 + 14,4} \right) + 2,1\\ = 2,5.\left( { - 1,4 + 14,4} \right) + 2,1\\ = 2,5.13 + 2,1\\ = 32,5 + 2,1\\ = 34,6\end{array}\)
2,
Cách 1:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 11,44 + 12,56 - 30,05 + 9\\ = \left( {11,44 + 12,56} \right) + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)
Cách 2:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 4.(2,86+3,14) - 30,05 + 9\\ = 4.6 + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)
a) \(2^5\cdot2^7\)
\(=2^{5+7}\)
\(=2^{12}\)
b) \(2^3\cdot2^2\)
\(=2^{3+2}\)
\(=2^5\)
c) \(2^4\cdot2^3\cdot2^5\)
\(=2^{4+3+5}\)
\(=2^{12}\)
d) \(2^2\cdot2^4\cdot2^6\cdot2\)
\(=2^{2+4+6+1}\)
\(=2^{13}\)
e) \(2\cdot2^3\cdot2^7\cdot2^4\)
\(=2^{1+3+7+4}\)
\(=2^{15}\)
f) \(3^8\cdot3^7\)
\(=3^{8+7}\)
\(=3^{15}\)
g) \(3^2\cdot3\)
\(=3^{2+1}\)
\(=3^3\)
h) \(3^4\cdot3^2\cdot3\)
\(=3^{4+2+1}\)
\(=3^7\)
I) \(3\cdot3^5\cdot3^4\cdot3^2\)
\(=3^{1+5+4+2}\)
\(=3^{12}\)
Lời giải chi tiết
12 1 13 12 – 02 (0 + 1)2 02 +12
22 1 + 3 23 32 – 12 (1 + 2)2 12 + 22
32 1 + 3 + 5 33 62 – 32 (2 + 3)2 22 + 32
43 102 – 62
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a) 2A = 2 + 2^2 + 2^3 +...+ 2^11
2A-A = (2 + 2^2 + 2^3 +...+ 2^11) - (1 + 2 + 2^2 +...+ 2^10)
A = 2^11 - 1
b) 3B = 3 + 3^2 + 3^3 +...+ 3^101
3B-B = (3 + 3^2 + 3^3 +...+ 3^101) - (1 + 3 + 3^2 +...+ 3^100)
2B = 3^101 - 3
B = \(\frac{\text{3^101 - 3}}{2}\)
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