Tính:
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
Chứng minh rằng với \(n\in N\)* thì:
a, \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
b, \(1^3+2^3+3^3+...+n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2\)
c, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)
Cho: A=\(1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)
Chứng tỏ A không phải là số nguyên
Các bn giúp mk vs,tí nữa là phải đi hx rùi.1h15' mk quay lại.Nhanh nha
Tìm n:
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
d) \(\left(0,7\right)^{3n+1}=10^3:7^3\)
Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
Bài 1:
a)Tìm x:
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
b) cho: \(M=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}\)
Bài 2:
a) Tìm x,y biết:
\(\frac{x+2y}{18}=\frac{1+4y}{24}=\frac{1+x+6y}{6x}\)
b) tìm ssos nguyên n để A mang giá trị nguyên và tính giá trị đó
\(A=\frac{9+3n}{n-4}\)
Tính :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{n+1}\right)vớin\in N\)
tính
a) \(3^{-2}.\left[\left(\frac{2}{3}\right)^{-4}\right].\left[\left(-1\frac{1}{2}\right)^{-3}\right]\)
b) \(\left[\left(0.02\right)^{-3}\right].10^{-4}.\left(\frac{4}{5}\right)^{-2}\)
c) \(\left[2^{-2}-\frac{3}{4}^{-4}.\left(\frac{-1}{2}^2\right)\right]:\left(10^{-1}+1\right)\)
Tính :
A = \(\left(1-\frac{2}{3}+\frac{4}{3}\right)-\left(\frac{4}{5}-1\right)+\left(\frac{7}{5}+2\right)\)
B = \(\left(-3+\frac{3}{4}-\frac{1}{3}\right):\left(5+\frac{2}{5}-\frac{2}{3}\right)\)
C = \(\left(\frac{3}{5}-\frac{4}{5}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right).\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1-\frac{11}{12}\right)\)
