Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

\(C=\left|-3\left(\dfrac{-13}{15}-\dfrac{17}{21}\right)\right|-\left|\dfrac{-13}{15}+\dfrac{17}{7}\right|+\left(-12+\dfrac{35}{3}\right):\left|-\dfrac{7}{6}\right|\\ =\left|-3.-\dfrac{176}{105}\right|-\left|-\dfrac{6}{35}\right|+\left(-\dfrac{1}{3}\right):\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{1}{3}:\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{2}{7}\\ =\dfrac{170}{35}-\dfrac{2}{7}=\dfrac{32}{7}.\)

tính giúp mình với mình đang cần gấp

cảm ơn.

A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)

= \(\dfrac{1}{2}\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)

= \(\dfrac{1}{2}\left(\sqrt{99}-\sqrt{3}\right)\)

B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35

= 33 + 2 + 333 + 2 + 3333 + 2 + ... + 333...33 + 2

= 2 . 99 + (33 + 333 + 3333 + ... + 333...3)

= 198 + \(\dfrac{1}{3}\)(99 + 999 + 9999 + ... + 999...99)

= 198 + \(\dfrac{1}{3}\)(10² - 1 + 10³ - 1 + 10⁴ - 1 + ... + 10¹⁰⁰ - 1)

= \(\left(\dfrac{10^{101}-10^2}{27}\right)+165\)

Ta có:

\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)

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

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100}{2}\)

\(=50\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\) \(=\dfrac{5}{39}\)

\(\dfrac{5}{39}\)

=\(\dfrac{5}{39}\)

\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}=\dfrac{13-3}{39}=\dfrac{10}{39}\)

\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)