Sửa đề: \(5^9\cdot49^2\)

\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)

\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)

=-34/243

1.  25 : 5,7 = 250/57

2.  30:2.8.4 = 480

3.  20:2^2.14= 70

4.  125:5^3.170= 170

5.  64:2^5.30.4=240

6.  (25:5^2.30): 15.7=14

       bạn à! Nhiều quá mình ko làm hết được. sorry nha.^-^                     

dù dài quá nhưng cố  gắng giúp mình làm hết nha

a, \(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+...+\(\dfrac{5}{99.101}\)

= 5.\(\dfrac{1}{1.3}\)+5.\(\dfrac{1}{3.5}\)+5.\(\dfrac{1}{5.7}\)+...+5.\(\dfrac{1}{99.101}\)

=5.(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{99.101}\))

=5.(\(\dfrac{2}{2}\).\(\dfrac{1}{1.3}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{3.5}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{5.7}\)+...+\(\dfrac{2}{2}\).\(\dfrac{1}{99.101}\))

=\(\dfrac{5}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\))

=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))

=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{101}\))

=\(\dfrac{5}{2}\).\(\dfrac{100}{101}\)

=\(\dfrac{250}{101}\)

=\(2\dfrac{48}{101}\)

b,\(\dfrac{-11}{23}\).\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\).\(\dfrac{-11}{23}\)-\(\dfrac{1}{23}\)

=\(\dfrac{-11}{23}\).(\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\))-\(\dfrac{1}{23}\)

=\(\dfrac{-11}{23}\).2-\(\dfrac{1}{23}\)

=\(\dfrac{-22}{23}\)-\(\dfrac{1}{23}\)

=-1

c,\(\dfrac{2.3}{7}\)+(\(\dfrac{2}{9}\)-\(1\dfrac{1}{3}\))-\(\dfrac{5}{3}\):\(\dfrac{1}{9}\)

=\(\dfrac{6}{7}\)+(\(\dfrac{2}{9}\)-\(\dfrac{4}{3}\))-\(\dfrac{5}{3}\).9

=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-\(\dfrac{5}{3}\).9

=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-15

=\(\dfrac{54}{63}\)-\(\dfrac{70}{63}\)-\(\dfrac{945}{63}\)

=\(\dfrac{-961}{63}\)=\(-15\dfrac{16}{63}\)

d,(20+\(9\dfrac{1}{4}\)):\(2\dfrac{1}{4}\)

=(20+\(\dfrac{37}{4}\)):\(\dfrac{9}{4}\)

=20:\(\dfrac{9}{4}\)+\(\dfrac{37}{4}\):\(\dfrac{9}{4}\)

=20.\(\dfrac{4}{9}\)+\(\dfrac{37}{4}\).\(\dfrac{4}{9}\)

=\(\dfrac{80}{9}\)+\(\dfrac{37}{9}\)

=\(\dfrac{117}{9}\)

Mình làm nhiều như thế thì sẽ không tránh được lỗi sai nên mình mong bạn hãy xem lại thật kĩ nhé!

bài 1.

A) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}=\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)=\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)

B)\(-\dfrac{5}{13}+\left(\dfrac{3}{5}+\dfrac{3}{1}-\dfrac{4}{10}\right)=-\dfrac{5}{13}+3+\left(\dfrac{3}{5}-\dfrac{4}{10}\right)=-\dfrac{5+3.13}{13}+\dfrac{1}{5}=-\dfrac{207}{65}\)C)\(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{-17}+\dfrac{-3}{15}=\left(\dfrac{5}{17}+\dfrac{2}{17}\right)+\left(\dfrac{-9}{15}+\dfrac{-3}{15}\right)=\dfrac{7}{17}-\dfrac{12}{15}=-\dfrac{33}{85}\)D)\(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{-1}{9}=\left(\dfrac{1}{9}+\dfrac{-1}{9}\right)+\left(\dfrac{3}{6}+\dfrac{1}{2}\right)-\dfrac{12}{17}=1-\dfrac{12}{17}=\dfrac{5}{17}\)

Giúp mình với

nhanh lên đi

bài 5:

A)\(\dfrac{1}{3}+\dfrac{-1}{4}+\dfrac{1}{5}+\dfrac{1}{-6}+\dfrac{-1}{-7}+\dfrac{1}{6}+\dfrac{-1}{5}+\dfrac{1}{4}+\dfrac{1}{3}\\ =\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{-1}{4}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{-1}{5}+\dfrac{1}{6}+\dfrac{1}{-6}+\dfrac{1}{7}\\ =\dfrac{2}{3}+\dfrac{1}{7}=\dfrac{17}{21}\)

B)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)