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A = 32 - 35 + 38 - 311 + ...... + 398 - 3101
=> 3A = 35 - 38 + 311 - 314 + ...... + 3101 - 3104
=> 3A + A = (35 - 38 + 311 - 314 + ...... + 3101 - 3104) + (32 - 35 + 38 - 311 + ...... + 398 - 3101)
=> 4A = -3104 + 32
=> 28A = 7(32 - 3104)
=> B + 28A = 3104 + 7(32 - 3104)
=> B + 28A = 7.32 - 6.3104 = 32 (7 - 2.3103)
(DẤU "." LÀ NHÂN NHA BẠN☺)
a, 13 + 23 + 33 + .... + 993 + 1003
= ( 1 + 2 + 3 +... + 99 + 100 )2
= 50502
c, 13 + 63 + 113 + ...... + 1013
= ( 1 + 6 + 11 + ... + 101 )2
= 10712
b, Sai đề không chia hết
Ta có:
A=3 +32 +33 +34 + 35 +...+39
A=(3+32+33) + (34+35+36) + (37+38+39)
A= 39 + 39. 34 + 39. 37
A= 39. (1+34+37)\(⋮\)39
Vậy A\(⋮\)39
A=3 +32 +33 +34 + 35 +...+39
A=(3+32+33) + (34+35+36) + (37+38+39)
A= 39 + 39. 34 + 39. 37
A= 39. (1+34+37)\(⋮\)39
Vậy A\(⋮\)39
\(A=1+6+6^2+...+6^{100}\)
\(6A=6+6^2+6^3+...+6^{101}\)
\(6A-A=\left(6+6^2+...+6^{101}\right)-\left(1+6+...+6^{100}\right)\)
\(5A=6^{101}-1\)
\(A=\frac{6^{101}-1}{5}\)
Hoàn toàn tương tự với các câu b) c)
\(A=1+6+6^2+6^3+...+6^{100}\)
\(6A=6+6^2+6^3+6^4+...+6^{101}\)
\(6A-A=\left(6+6^2+6^3+6^4+...+6^{101}\right)-\left(1+6+6^2+...+6^{100}\right)\)
\(5A=6^{101}-1\)
\(A=\frac{6^{101}-1}{5}\)
\(\left(125^{100}.2^{160}\right):\left(5^{289}.4^{80}\right)\)\(=\frac{\left(5^3\right)^{100}.2^{100}}{5^{289}.\left(2^2\right)^{80}}\)\(=\frac{5^{300}.2^{100}}{5^{289}.2^{160}}\)\(=\frac{5^{11}}{2^{60}}\)
\(\left(9^8.5^8\right):\left(3^7.27^3.5^4\right)\)\(=\frac{\left(3^2\right)^8.5^8}{3^7.\left(3^3\right)^3.5^4}=\frac{3^{16}.5^4}{3^7.3^9}=\frac{3^{16}.5^4}{3^{16}}=5^4=625\)
\(\left(1024.27^8\right):\left(2^9.3^{23}\right)=\frac{2^{10}.\left(3^3\right)^8}{2^9.3^{23}}=\frac{2.3^{24}}{3^{23}}=2.3=6\)
\(\left(625.2^7+25^2.64\right):\left(2^6.5^4.3\right)=\frac{25^2.128+25^2.64}{2^6.\left(5^2\right)^2.3}=\frac{25^2.\left(128+64\right)}{64.25^2.3}=\frac{192}{192}=1\)
\(A=3^2-3^5+3^8-3^{11}+...-3^{101}\)
\(\Rightarrow3A=3^5-3^8+3^{11}-3^{14}+...-3^{104}\)
\(\Rightarrow3A+A=\left(3^5-3^8+3^{11}-3^{14}+...-3^{104}\right)+\left(3^2-3^5+3^8-3^{11}+...-3^{101}\right)\)
\(\Rightarrow4A=-3^{104}+3^2\)
\(\Rightarrow28A=7\left(3^2-3^{104}\right)\)
\(\Rightarrow B+28A=3^{104}+7\left(3^2-3^{104}\right)\)
\(\Rightarrow B+28A=7.3^2-6.3^{104}=3^2\left(7-2.3^{103}\right)\)