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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{9991}{9992}+\frac{1}{9}-\frac{2}{19984}-\frac{5}{495}\)
\(=0,99989991993+0,11111111111-0,00010008006-0,0101010101\)
\(=1,10080994088\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
Vậy \(P=-\frac{97}{99}\)
P=-1/1.2-1/2.3-...-1/98.99-1/99
P=-(1/1.2+1/2.3+...+1/98.99+1/99)
P=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
= \(\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
= \(\frac{4}{5}\)
\(\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}=\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}=\frac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\frac{5}{3}.\frac{3}{7}+\frac{5}{3}.\frac{5}{7}-\frac{5}{3}\)
=\(\frac{5}{3}.\left(\frac{3}{7}+\frac{5}{7}\right)-\frac{5}{3}\)
= \(\frac{5}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
\(2\frac{5}{12}-99\frac{213}{9115}:\frac{12}{5}+98\frac{213}{9115}:\frac{12}{5}\)
\(=2\frac{5}{12}-\left(99\frac{213}{9115}+98\frac{214}{9115}\right):\frac{12}{5}\)
\(=2\frac{5}{12}-197:\frac{12}{5}\)
\(=2\frac{5}{12}-197.\frac{5}{12}\)
\(=2\frac{5}{12}-197\frac{5}{12}\)
\(=\left(2-197\right)\frac{5}{12}\)
\(=-195\frac{5}{12}\)
\(=-2335\)
~ Chắc đúng ~
# Chúc bạn học tốt #
Bài làm
Mik sai mất rồi. Mik sửa nè:
\(=-195\frac{5}{12}\)
\(=-\frac{2335}{12}\)
~ Thế mới đúng nha ~
# Chúc bạn học tốt #
Rút gọn phân số rồi tính như thường :)
\(-\frac{9991}{9992}+\frac{1}{99}-\frac{2}{19984}-\frac{5}{495}\)
\(=-\frac{9991}{9992}+\frac{1}{99}-\frac{1}{9992}-\frac{1}{99}\)
\(=\left(-\frac{9991}{9992}-\frac{1}{9992}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)
\(=-1+0\)
\(=-1\)