Question

[Đỗ Thuỳ Linh]

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

Answer 1

[Nguyễn Huy Tú]

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{3004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(\Rightarrow P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)

\(\Rightarrow P=\frac{3}{15}-\frac{10}{15}\)

\(\Rightarrow P=\frac{-7}{15}\)

Vậy \(P=\frac{-7}{15}\)

Answer 2

[Đỗ Thuỳ Linh]

Câu còn lại ko làm được hả bạn