Giải:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{2015.2018}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=\dfrac{42}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=0,977240464-\dfrac{1}{2018}\)

\(\Leftrightarrow A=0,9767449238\approx0,98\)

Vậy ...

\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)

\(E=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{31.34}\right)\)

\(E=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)

\(E=\frac{2}{3}\left(1-\frac{1}{34}\right)\)

\(E=\frac{2}{3}.\frac{33}{34}\)

\(E=\frac{11}{17}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

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

tại sao ko đóng mở ngoặc phép tính rồi nhân 3 vậy?

Ta có :

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=\)\(1-\frac{1}{43}\)

\(=\)\(\frac{42}{43}\)

\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)

\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)

\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)

Làm từng phần nha bạn

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)

Đặt \(A+x=\frac{299}{301}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)

\(A=1-\frac{1}{301}\)

\(A=\frac{300}{301}\)

=> \(\frac{300}{301}+x=\frac{299}{301}\)

\(x=\frac{299-300}{301}\)

\(x=-\frac{1}{301}\)

\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)

\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=\frac{303}{304}\)

\(A=\frac{505}{304}\)

Trả lời

3/1.4+3/4.7+3/7.10

=3/1.3/10

=3/10

Chúc bạn học tốt #

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}\)

\(=\frac{9}{10}\)

1.

`16 + (27 - 7.6 ) - (94 -7 - 27.99)`

`= 16+ 27 - 7.6 - 94 + 7 + 27.99`

`= 16 + 27(99 +1) - 7(6-1) - 94`

`= -78 + 27.100 - 7.5`

`= 2587`

2.

`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`

`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`

`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`

`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`

`3/2A = 1 - 1/100`

`3/2 A= 99/100`

`A= 99/100 : 3/2`

`A=33/50`

Vậy `A= 33/50`

1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99

                                           =(27+27.99)+(27+7-94)+16

                                           =27.100-60+16

                                           =2700-44=2656

2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)

     =\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

     =\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

1) \(16+\left(27-7.6\right)-\left(94-7-27.99\right)\)

=\(16+27-7.6-94+7+27.99\)

=\(\left(27+27.99\right)+\left(-7.6+7\right)+\left(16-94\right)\)

=\(27\left(1+99\right)+7\left(-6+1\right)-78\)

=\(27.100-7.5-78=2700-35-78=2587\).

2) \(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)

\(A=\dfrac{2.3}{1.4.3}+\dfrac{2.3}{4.7.3}+\dfrac{2.3}{7.10.3}+...+\dfrac{2.3}{97.100.3}\)

\(A=\dfrac{2}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=\dfrac{2}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)=\dfrac{2}{3}.\dfrac{99}{100}=\dfrac{33}{50}\)