Tinh:1/2002+(2003.2001/2002)-2003
Tinh :\(\frac{1}{2002}\)+\(\frac{2003.2001}{2002}\)-2003
\(\frac{1}{2002}+\frac{2003.2001}{2002}-2003\)
\(\frac{1}{2002}+\frac{2003.2001}{2002}-2003\)
\(=\frac{1}{2002}+\frac{2003.3001}{2002}-\frac{2003.2002}{2002}\)
\(=\frac{1}{2002}+\frac{2003.2001-2003.2002}{2002}\)
\(=\frac{1}{2002}+\frac{-2003}{2002}=\frac{1+\left(-2003\right)}{2002}=\frac{-2002}{2002}=-1\)
Tinh nhanh
2002 × 2002 +1
2003 × 2002 - 2002
\(=\frac{2002x2002+1}{2002\left(2003-1\right)}=\frac{2002x2002+1}{2002x2002}=1+\frac{1}{2002x2002}\)
tinh nhanh 2001+2002+1981+2003*21/2003*2002-2001*2002
Bài 1: Thực hiện phép tính:
a) ( 2- \(\dfrac{3}{2}\)). ( 2- \(\dfrac{4}{3}\)). (2- \(\dfrac{5}{4}\)). ( 2- \(\dfrac{6}{5}\))
b) \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)
c)( \(\dfrac{2003}{2004}+\dfrac{2004}{2003}\)):\(\dfrac{8028045}{8028024}\)
d) 4+ \(\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{4}}}}\)
Tinh nhanh
2003×14+1998+2001×2002 / 2002+2002×503+504×2003
Viet ro loi giai nhe! Ai nhanh mk tk cho!
2003x14+1998+2001x2002/2002+2002x503+504x2003 =2003x14+1998+2001x1+2002x503+504x2003 =2003x14+1998+2001+2002x503+504x2003 =28042+1998+2001+1007006+1009512 =(28042+1009512+1007006)+1998+2001 =2044560+1998+2001=2048559 TK MIk nha bn! mk tra loi dau tien ma ! Giu loi hua nhe!
2003 x 14 + 1988 + 2001 x 2002
=28042 + 1988 + 4006002
=4036032
2002 + 2002 x 503 +504 x 2002
=2002 x 1 + 2002 x 503 + 504 x 2002
=2002 x (1 + 503 + 504)
=2002 x 908
=1817816
tính :
a, (2-\(\dfrac{3}{2}\) ) . (2-\(\dfrac{4}{3}\) ) . (2-\(\dfrac{5}{4}\) ) . (2-\(\dfrac{6}{5}\) )
b,\(\dfrac{1}{2002}\) + \(\dfrac{2003.2001}{2002}\) - 2003
c, (\(\dfrac{2003}{2004}\) + \(\dfrac{2004}{2003}\) ) : \(\dfrac{8028025}{8028024}\)
giúp mình với mình đang cần gấp
a, \(\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)
\(=\left(\dfrac{4}{2}-\dfrac{3}{2}\right)\left(\dfrac{6}{3}-\dfrac{4}{3}\right)\left(\dfrac{8}{4}-\dfrac{5}{4}\right)\left(\dfrac{10}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
b. \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)\(=\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-\dfrac{2003.2002}{2002}\) = \(\dfrac{1+2003.2001-2003.2002}{2002}\) = \(\dfrac{1+\left(2003\left(2001-2002\right)\right)}{2002}\) = \(\dfrac{1+2003.\left(-1\right)}{2002}\) = \(\dfrac{1+\left(-2003\right)}{2002}\) = \(\dfrac{-2002}{2002}=-1\)
Chúc nguyễn hồng nhung học tốt
tinh nhanh
(7^2003+7^2002):7^2002
(72003 + 72002) : 72002
= (72003 + 72002) . 1/72002
\(=\frac{7^{2003}}{7^{2002}}+\frac{7^{2002}}{7^{2002}}=7+1=8\)
= 7 ^ 2002 + 7 ^2003 ÷ 7 ^ 2002 +7 ^ 2002 = 7
Tìm 2 chữ số tận cùng của các tổng :
a,A=1^2002 + 2^2002+ 3 ^2002+........+2004^2002
b,B=1^2003 + 2^2003 + 3^2003+....+2004^2003