=>A=1/2.(2/1.3+4/3.7+6/7.13+...+20/91.111)

=>A=1/2.(3-1/1.3+7-3/3.7+13-7/7.13+...+111-91/91.111)

=>A=1/2.(1-1/3+1/3-1/7+1/7-1/13+...+1/91-1/111)

=>A=1/2.(1-1/111)

=>A=1/2.100/111

=>A=50/111

T*ck cho mìn nhóe!!!

-13.21-13.80+13

=-13.(21-80+1)

=-13.100

=-1300

-1+2-3+4-5+6-...-2021+2022

=(-1+2)+(-3+4)+...+(-2021+2022)

=1+1+1+...+1 (1011 số hạng)
=1011

-13 . 21 - 13 . 80 + 13

13 ( -21 - 80 + 1 )

13 . ( -100 )

- 1300

\(-13\cdot21-13\cdot80+13=13\left(-21-80+1\right)=13\cdot\left(-100\right)=-1300\)

a; C = \(\dfrac{3}{1.3}\) + \(\dfrac{3}{3.5}\) + \(\dfrac{3}{3.7}\) + ... + \(\dfrac{3}{49.51}\)

   C = \(\dfrac{3}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{49.51}\))

  C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{51}\))

 C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{51}\))

 C = \(\dfrac{3}{2}\).\(\dfrac{50}{51}\)

C = \(\dfrac{25}{17}\)

\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)

\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)

$A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}$

$A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}$

$A=\frac{1}{1}-\frac{1}{99}$

$A=\frac{98}{99}$

=

21​−51​+51​−71​+....+951​−981​

=12−198=21​−981​tự làm tiếp nha ( giống câu a)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{101.103}\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{101.103}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)

=\(\frac{1}{2}.\frac{102}{103}\)

=\(\frac{51}{103}\)

51/103

A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{101.103}\)

A = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{101.103}\))

A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{101}\) - \(\dfrac{1}{103}\))

A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{103}\))

A = \(\dfrac{1}{2}\). \(\dfrac{102}{103}\)

A = \(\dfrac{51}{103}\)

Em ơi thừa số thứ ba phải là \(\dfrac{1}{5.7}\) mới đúng em nhé.

\(32-2\left\{3^3-4\left[5^6:5^4-\left(3.7-1\right)\right]\right\}\)

\(=32-2\left[27-4\left(25-20\right)\right]\)

\(=32-2\left(27-100+80\right)\)

\(=32-2.7=32-14=18\)