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a = \(\frac{-8^3.6^4}{18^{12}}\)
= \(\frac{-2^9.2^4.3^4}{2^{12}.3^{24}}\)
= \(\frac{-2^{13}.3^4}{2^{12}.3^{24}}\)
= \(\frac{-2}{3^{20}}\)
Hk tốt
a)\(\frac{4.3}{9.32}=\frac{1}{3.8}=\frac{1}{24}\)
b)\(\frac{9.6-9.3}{18}=\frac{9.3}{9.2}=\frac{9}{1}=9\)
c)\(\frac{232323}{494949}=\frac{232323:10101}{494949:10101}=\frac{13}{49}\)
Rút gọn
A, \(\dfrac{125}{1000}\)
\(\dfrac{125}{1000}=\dfrac{125:125}{1000:125}=\dfrac{1}{8}\)
B, \(\dfrac{3600-75}{4800-175}\)
\(\dfrac{3600-75}{4800-175}=\dfrac{3525}{4625}\)
\(\Rightarrow\dfrac{3525}{4625}=\dfrac{3525:25}{4625:25}=\dfrac{141}{185}\)
C, \(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\)
\(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\dfrac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\dfrac{\left(-5\right)}{3^2}=\dfrac{-5}{9}\)
A = 1 + (1+ 1).2 + (1 + 2).3 + (1+3).4 + ...+ (1 + n-1). n
A = 1 + (2+1.2) + (3+ 2.3) + (4 + 3.4) + ....+ ( n + (n -1).n)
A = (1+ 2 + 3 + 4 + ...+ n) + (1.2 + 2.3 + 3.4 + .....+ (n-1).n)
Tính B = 1+ 2+ 3 + ...+ n = (n +1).n/ 2
C = 1.2+ 2.3 + 3.4 + ...+ (n-1).n
=> 3.C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ (n-1).n.3
3C = 1.2.3 + 2.3. (4 -1) + 3.4.(5 - 2) + ... + (n -1).n [(n+ 1) - (n -2)]
3C = [1.2.3 + 2.3.4 + ....+ (n-1).n.(n +1)] - (1.2.3 + 2.3.4 + ... + (n-2)(n -1).n)
3C = (n -1).n (n +1) => C = (n -1).n.(n +1)/ 3
Vậy A = (n +1).n/ 2 + (n -1).n(n +1)/3