\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)

=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)  

=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\) 

=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\) 

=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\) 

=\(-\dfrac{4}{7}-\dfrac{3}{10}\) 

=\(-\dfrac{40}{70}-\dfrac{21}{70}\)

=\(-\dfrac{61}{70}\)

   (3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))

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

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

= - 8  + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)

= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)

= -7 + 3 + \(\dfrac{5}{6}\)

= - 4 + \(\dfrac{5}{6}\)

= \(\dfrac{-19}{6}\)

\(\dfrac{2}{5}+\dfrac{3}{7}=\dfrac{14}{35}+\dfrac{15}{35}=\dfrac{29}{35}\)

\(\dfrac{1}{7}+\dfrac{3}{7}+\dfrac{4}{7}=\left(\dfrac{1}{7}+\dfrac{3}{7}\right)+\dfrac{4}{7}=\dfrac{1}{7}+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\left(\dfrac{1}{7}+\dfrac{4}{7}\right)+\dfrac{3}{7}\)

tính bằng cách thuân tiện nhé

Ôi rồi ôi

câu này thì chịu 

a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]

= -1/24 - [ 1/4 +3/8 ]

= -1/24 - 5/8

= -2/3.

a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]

= -1/24 - [ 1/4 +3/8 ]

= -1/24 - 5/8

= -2/3.

Xin lỗi nha chỉ có thời gian làm câu a) thôi

đến 1/5 là xuống phần mẫu số nhé

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

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

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

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

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

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

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

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

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

a) = 29/15

b) = 7/15

c) = 1

d) = 3

e) = 67/17

f) = 2

mk nhanh nhất tk cho mk nha

a/\(\frac{3}{5}+\frac{4}{3}=\frac{9}{15}+\frac{20}{15}=\frac{29}{15}\)

b/\(\frac{2}{3}-\frac{1}{5}=\frac{10}{15}-\frac{3}{15}=\frac{7}{15}\)

c/\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{7}{6}\)

d,\(\frac{3}{5}+\frac{4}{7}+\frac{7}{5}+\frac{3}{7}=\left(\frac{3}{5}+\frac{7}{5}\right)+\left(\frac{4}{7}+\frac{3}{7}\right)=2+1=3\)

a,  \(\frac{3}{5}+\frac{4}{3}=\frac{9}{15}+\frac{20}{15}=\frac{29}{15}\)

b,  \(\frac{2}{3}-\frac{1}{5}=\frac{10}{15}-\frac{3}{15}=\frac{7}{15}\)

c,  \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1\)

d,  \(\frac{3}{5}+\frac{4}{7}+\frac{7}{5}+\frac{3}{7}=\left(\frac{3}{5}+\frac{7}{5}\right)+\left(\frac{4}{7}+\frac{3}{7}\right)=\frac{10}{5}+\frac{7}{7}=2+1=3\)

e,  \(\frac{1}{5}+\frac{2}{9}+\frac{7}{9}+\frac{4}{5}+\frac{16}{17}+1=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{2}{9}+\frac{7}{9}\right)+\frac{16}{17}+1\)

 \(=1+1+1+\frac{16}{17}=3+\frac{16}{17}=\frac{3}{1}+\frac{16}{17}=\frac{51}{17}+\frac{16}{17}=\frac{67}{17}\)

f,  \(\frac{1}{3}\times\frac{2}{7}+\frac{1}{3}\times\frac{5}{7}+\frac{15}{9}=\left(\frac{1}{3}\times\frac{2}{7}\right)+\left(\frac{1}{3}\times\frac{5}{7}\right)+\frac{15}{9}=\frac{2}{21}+\frac{5}{21}+\frac{15}{9}\)

\(=\frac{7}{21}+\frac{15}{9}=\frac{63}{189}+\frac{315}{189}=\frac{378}{189}=2\)

Ta có

A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)

Đặt C = 1 + 7 + 7² + 7³+...+7¹¹

7C = 7 + 7² + 7³ + ... + 7¹¹ + 7¹²

=> 6C = 7¹² - 1

C = \(\dfrac{7^{12}-1}{6}\)

Đặt D = 1 + 7 + 7² + 7³+...+7¹⁰

7D = 7 + 7² + 7³ + ... + 7¹⁰ + 7¹¹

=> 6D = \(7^{11}-1\)

D = \(\dfrac{7^{11}-1}{6}\)

=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)

A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)

A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)

A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003

Lại có:

B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\

Đặt H = \(1+3+3^2+3^3+...+3^{11}\)

3H = \(3+3^2+3^3+...+3^{12}\)

=> 2H = \(3^{12}-1\)

H = \(\dfrac{3^{12}-1}{2}\)

Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)

3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)

=> 2Q = \(3^{11}-1\)

Q = \(\dfrac{3^{11}-1}{2}\)

=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)

B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)

B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)

B = \(\dfrac{3^{12}-1}{3^{11}-1}\)

B = 3, 00001129

Vì 7, 000000003 > 3, 00001129

=> A > B

Vậy A > B