\(M=1+2+2^2+...+2^{50}\)
\(\Rightarrow2M=2.\left(1+2+2^2+...+2^{50}\right)\)
\(2M=2+2^2+2^3+...+2^{51}\)
\(\Rightarrow2M-M=\left(2+2^2+2^3+...+2^{51}\right)-\left(1+2+2^2+...+2^{50}\right)\)
\(\Rightarrow M=2^{51}-1<2^{51}=N\)
Vậy M < N.
\(M=1+2+2^2+...+2^{50}\)
\(\Rightarrow2M=2.\left(1+2+2^2+...+2^{50}\right)\)
\(2M=2+2^2+2^3+...+2^{51}\)
\(\Rightarrow2M-M=\left(2+2^2+2^3+...+2^{51}\right)-\left(1+2+2^2+...+2^{50}\right)\)
\(\Rightarrow M=2^{51}-1<2^{51}=N\)
Vậy M < N.
so sanh M va N, biet:
N=364
M=2(3+1)(32+1)(33+1)(316+1)(332+1)
cho m va n la 2 so tu nhien va p la mot so nghuyen thoa man 7/m-1=m+n/p
c/m p 2 =n+2
cho m va n la 2 so tu nhien va p la mot so nghuyen thoa man 7/m-1=m+n/p
c/m p 2 =n+2
cho 2 stn lẻ m va n nguyen to cung nhau t/m m^2+2 chia het cho n va n^2+2 chia het cho m
Bai 1 so sanh cac so sau :
2^0+2^1+2^2+2^3+......+2^50 va 2^51
Neu ai gia dc bai toan nay thi tui se tich cho
so sanh (2/3)^45 va (2/3)^50
so sanh A va B
biết A=1/2!+2/3!+3/4!+.......+2015/2016!
và B=1,02016
biết n!=1*2*3*......*n
lưu ý * là dấu nhân
cho tam giac abc vuong tai a goi o la giao diem cac duong phan giac bd va ce lay diem m va n thuoc bc sao cho ba=bm ca=cn cm oa^2=1/2 mn^2
Cho A= 1+2+2^2+2^3+.....+2^10. So sanh A va 2^11