Bài 2: Cho A= 1^3 + 2^3 + 3^3 +...+ 10^3
tính B= 2^3 + 4^3 + 6^3 +...+ 20^3 theo A
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a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
\(3\div\frac{1}{2}=3\times\frac{2}{1}=6\)
\(\frac{1}{2}\div3=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\)
2.Tính theo mẫu :
b) \(\frac{6}{25}\div\frac{21}{20}=\frac{6}{25}\times\frac{20}{21}=\frac{6\times20}{25\times21}=\frac{2\times3\times4\times5}{5\times5\times3\times7}=\frac{8}{35}\)
\(\frac{40}{7}\times\frac{14}{5}=\frac{40\times14}{7\times5}=\frac{5\times8\times7\times2}{7\times5}=16\)
\(\frac{17}{13}\div\frac{51}{26}=\frac{17}{13}\times\frac{26}{51}=\frac{17\times13\times2}{13\times17\times3}=\frac{2}{3}\)
Chúc bạn hok tốt !
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
\(B=2^3+4^3+6^3+...+20^3\\=\left(2.1\right)^3+\left(2.2\right)^3+\left(2.3\right)^3+...+\left(2.10\right)^3\\ =2^3.1^3+2^3.2^3+2^3.3^3+...+2^3.10^3\\ =2^3\left(1^3+2^3+3^3+...+10^3\right)\\ =8A\)