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\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)-2\times\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(A=\frac{1}{1002}+\frac{1}{1003}+...\frac{1}{2002}\)= B
=> A/ B = 1
\(\frac{2003\times4+1998+2001\times2002}{2002+2002\times1002+2002\times1003}\)
\(=\frac{2003\times4+2\times999+2001\times2\times1001}{2002.\left(1+1002+1003\right)}\)
\(=\frac{2\times\left(2003\times2+999+2001\times1001\right)}{1001\times2\times\left(1+1002+1003\right)}\)
\(=\frac{2003\times2+999+2001\times1001}{1001\times\left(1+1002+1003\right)}\)
\(=1\)
mk ko bít
a ) \(\frac{2003\times14+1988+2001+2002}{2002+2002\times503+504\times2002}\)
= \(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
= \(\frac{2002\times14+14+1998+2001\times2002}{2002\times1008}\)
= \(\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
= \(\frac{2002\times\left(14+1+2001\right)}{2002\times1008}\)
= \(\frac{2016}{1008}\)
= 2
b ) Đặt A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=> 2A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
=> 2A - A = ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 ) - ( 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 )
=> A = 1/2 - 1/128
A = 63/128
a) 2003x14+1988+2001+2002 / 2002+2002x503+504x2002
=(2002+1)x14+1988+2001+2002 / 2002x(503+1+504)
=2002x14+(14+1988)+2002+2001 / 2002x1008
=2002x(14+1+1)+2001 / 2002x1008
đến đoạn nay mk thấy đề có ì đó sai sai rồi đó, 2001 đáng lẽ phải bằng 2002 mới đúng chứ, đề ko lỗi thì cho mk xin lỗi nha
b) 1/4 + 1/8 + 1/16 + 1/64 + 1/128 (bạn thiếu 100 ở đầu mẫu)
gọi tổng sau là a, ta có
A = 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7
2xA = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6
2xA-A = (1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6) - (1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7)
A = 1/2 - 1/2^7
A = 2^6-1/2^7
chúc bạn học tốt nha
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(A=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2001}+\frac{1}{2002}=B\)
=> A/B = 1