\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)

\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)

\(B=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(B=100\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(B=100\left(1-\frac{1}{100}\right)\)

\(B=100.\frac{99}{100}=99\)

Đúng òi nên xin chúc mừng nha !!!

a, \(\dfrac{3}{5}\). (\(-\dfrac{9}{11}\)) + \(\dfrac{-3}{5}\).\(\dfrac{2}{11}\) + \(\dfrac{3}{5}\)

= \(\dfrac{3}{5}\).( - \(\dfrac{9}{11}\) - \(\dfrac{2}{11}\) + 1)

= \(\dfrac{3}{5}\).(- \(\dfrac{11}{11}\) + 1)

= \(\dfrac{3}{5}\).(1-1)

= \(\dfrac{3}{5}\).0

= 0 

\(A=\frac{92-\frac{1}{9}-\frac{2}{10}-...-\frac{91}{99}-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+...+\frac{1}{495}+\frac{1}{500}}\)

Đặt:  \(M=92-\frac{1}{9}-\frac{2}{10}-...-\frac{91}{99}-\frac{92}{100}\)

Tách 92  thành tổng của 92 số 1.

\(M=1-\frac{1}{9}+1-\frac{2}{10}+...+1-\frac{91}{99}+1-\frac{92}{100}\)

\(M=\frac{8}{9}+\frac{8}{10}+...+\frac{8}{99}+\frac{8}{100}\)

\(M=\frac{40}{45}+\frac{40}{50}+...+\frac{40}{495}+\frac{40}{500}\)

Thay M vào A:

\(\Rightarrow A=\frac{\frac{40}{45}+\frac{40}{50}+...+\frac{40}{495}+\frac{40}{500}}{\frac{1}{45}+\frac{1}{50}+...+\frac{1}{495}+\frac{1}{500}}\)

\(\Rightarrow A=\frac{40\cdot\left(\frac{1}{45}+\frac{1}{50}+...+\frac{1}{495}+\frac{1}{500}\right)}{\left(\frac{1}{45}+\frac{1}{50}+...+\frac{1}{495}+\frac{1}{500}\right)}\)

\(\Rightarrow A=40\)

PP/ss: Tớ ko chắc đâu :)))

cảm ơn bạn  nhìu lắm

\(\Rightarrow A=\frac{600}{12}+\frac{240}{12}+\frac{100}{12}+\frac{60}{12}+\frac{40}{12}.\)

 \(=\frac{2080}{12}=\frac{520}{3}.\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{7}\right)+.....+\frac{1}{2}\left(\frac{50}{89}-\frac{50}{99}\right)+\frac{1}{99}\)

\(=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{7}+\frac{50}{7}-\frac{50}{8}+.................+\frac{50}{98}-\frac{50}{99}\right)+\frac{1}{99}\)

 \(=\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)+\frac{1}{99}\)

Từ A và C ta có \(B=\frac{520}{3}+\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)+\frac{1}{99}\)

                             \(=\left(\frac{520}{3}+\frac{1}{99}\right)+\frac{1}{2}\left(\frac{50}{6}-\frac{50}{99}\right)\)

                               \(=\frac{17161}{99}+\frac{1}{2}x\frac{775}{99}\)

                                \(=\frac{17161}{198}+\frac{17161}{99}=\frac{17161}{198}+\frac{34322}{198}=\frac{17161}{66}\)

vậy biểu thức\(B=\frac{17161}{66}\)

Ta chia B thành 2 phần là A và C

Ta có :\(A=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}\)

           \(B=\frac{100}{6x7}+..........+\frac{100}{98x99}+\frac{1}{99}\)

TỪ NHA BN MK LÀM ĐÉN ĐÓ TÍ NỮA MK LÀM TIẾP H MK CÓ VIỆC BN ZÙI