Do -2002/2003>-1 và -2005/2004<-1
\(\Rightarrow\)-2002/2003>-2005/2004
Do -2002/2003>-1 và -2005/2004<-1
\(\Rightarrow\)-2002/2003>-2005/2004
Tính :
P=\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
Q=\(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\right)+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{3}{12}}:\frac{1890}{2005}+115\)
Giúp mình với mình đang cần gấp, thanks các bạn nhìu
1/ So sánh các số hữu tỉ sau
a/ \(\frac{13}{17}và\frac{46}{50}\)
b/ \(\frac{33}{131}và\frac{53}{217}\)
c/ \(\frac{41}{91}và\frac{411}{911}\)
d/ \(\frac{2001}{2002}và\frac{2005}{2003}\)
e/ \(\frac{-2005}{2010}và\frac{2001}{2002}\)
Tìm x biết: \(\frac{X-1}{2004}+\frac{X-2}{2003}-\frac{X-3}{2002}=\frac{X-4}{2001}\)
cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)CMR 4(a-b)(b-c)=\(\left(c-a\right)^2\)
Cho \(\frac{a+b}{2002}\)=\(\frac{b+c}{2003}\)=\(\frac{c+a}{2004}\)
Chứng minh rằng:(a-c)(b-a)=\(^{c-b^2}\)
\(\frac{1}{2002}+\frac{2003.2001}{2002}-2003\)
Cho \(\left(2x_1-3y_1\right)^{2004}+\left(2x_2-3y_2\right)^{2004}+...+\left(2x_{2015}-3y_{2015}\right)^{2004}\)
C/M rằng \(\frac{x_1+x_2+x_3...+x_{2004}+x_{2005}}{y_1+y_2+y_3+...+y_{2005}+y_{2005}}=\frac{3}{2}\)
So sánh
\(\left(\frac{9}{11}-0,81\right)^{2005}\) và \(\frac{1}{10^{4010}}\)
\(\frac{3}{5}< \frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+.....+\frac{1}{4006}< \frac{3}{4}\)