2004x2005+2006x6-6
2005x1997+4x2005
2004x2005+2006x6-6/2005x1997+4x2005
\(\dfrac{2004\times2005+2006\times6-6}{2005\times1997+4\times2005}\\ =\dfrac{2004\times2005+\left(2005+1\right)\times6-6}{2005\times\left(1997+4\right)}\\ =\dfrac{2004\times2005+2005\times6+6-6}{2005\times2001}\\ =\dfrac{2005\times\left(2004+6\right)}{2005\times2001}\\ =\dfrac{2010}{2001}\\ =\dfrac{670}{667}\)
Tính nhanh:
a.2,3+3,4+4,5+...+11,1+12,2
b.2004x2005+2006x6-6 / 2005x197+4x2005
2004x2005+2006x6-6/ 2005 x 1997 +4 x2005
bằng 48167947570 và đừng đăng câu hỏi linh tinh lên olm .vn mog bạn CHÚ Ý
tôi đăng linh tinh j
A.\(\frac{2004x2005+2006x6-6}{2005x1997+4x2005}\) \(\frac{1999x2000+2001x5-5}{504x2000+500x2000}\)
B.\(\frac{2003x4+1998+2001x2002}{2002+2002x502+500x2002}\) \(\frac{2000x4+1995+2001x1995}{1995x495+1995x5+1995x3}\)
C.\(\frac{72:2x574+286x2x64}{4+4+8+12+20+...+220}\) \(\frac{72+36x2+24x3+18x4+12x6+168}{2+2+4+6+...+512+1024}\)
đề bài : so sánh
2003x2004-1/2003x2004 với 2004x2005-1/2004x2005
ta có: \(\frac{2003\times2004-1}{2003\times2004}=\frac{2003\times2004}{2003\times2004}-\frac{1}{2003\times2004}=1-\frac{1}{2003\times2004}\)
\(\frac{2004\times2005-1}{2004\times2005}=\frac{2004\times2005}{2004\times2005}-\frac{1}{2004\times2005}=1-\frac{1}{2004\times2005}\)
ta có: \(\frac{1}{2003\times2004}>\frac{1}{2004\times2005}\Rightarrow1-\frac{1}{2003\times2004}<1-\frac{1}{2004\times2005}\)
\(\frac{2003\times2004-1}{2003\times2004}<\frac{2004\times2005-1}{2004\times2005}\)
So sánh :
A= 2003x2004-1 / 2003x2004 và B=2004x2005-1 / 2004x2005
A=2003x2004-1/2003x2004
B=2004x2005-1/2004x2005
A= 1-2003x2004-1/2003x2004=1/2003x2004
B=1-2004x2005-1/2004x2005=1/2004x2005
Vì 1/2003x2004<1/2004x2005 => A>B.
K nhé
So Sánh :
2003x2004-1/2003x2004 và 2004x2005-1/2004x2005
Trình bày rõ ràng nha các bạn
hãy so sánh ps \(\frac{2003x2004-1}{2003x2004}và\frac{2004x2005-1}{2004x2005}\)
bài so sánh :
\(\frac{2003x2004-1}{2003x2004}\)và \(\frac{2004x2005-1}{2004x2005}\)
\(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)
\(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)
Vì \(\frac{1}{2003.2004}>\frac{1}{2004.2005}\)
=> \(1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
=> \(\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)