b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

A = ( 1 - 1/2) . ( 1 - 1/3 ) . (1-1/4) ....(1-1/2015) . (1-1/2016)

A= 1/2 . 2/3 . 3/4...2014/2015 . 2015/2016

A = 1 . 2 . 3 . 4 ... 2014 . 2015/ 2 . 3 . 4 ... 2015 . 2016

A = 1/ 2016

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\frac{2}{9}\)

\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)

\(x-2=8\)

\(x=8+2\)

\(x=10\)

Vậy \(x=10\)

\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)

2(x-2)=16

x-2=8

x=10
 

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x+2}{42} +\frac{x+2}{56}+\frac{x+2}{72}=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-2\right)\frac{4}{9}=\frac{16}{9}\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

Quên . Phải đóng ngoặc mấy chỗ x-2

\(\dfrac{x-2}{12}+\dfrac{x-2}{20}\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(x-2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\)

\(x-2.\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(x-2=\dfrac{16}{9}:\dfrac{2}{9}\)

\(x-2=8\)

x=10

Vậy x bằng 10 thì thoả mãn đề bài

tich nhé đúng đấy

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

giờ thì tự làm nha

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

giờ thì tự làm nha

Ai k mk mk k lại

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)

\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)

\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{11}{6}\)

ớ ko có vế phải seo tính đc

à quên =10 chờ tí nhé

\(\frac{x-2}{3.4}+\frac{x-2}{4.5}+...+\frac{x-2}{8.9}=10\)

tới đây tự nghĩ là ra

x thuộc Z

\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)

\(x-2=\frac{32}{91}\)

\(x=\frac{32}{91}+2\)

\(x=\frac{212}{91}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

=1-1/9

=8/9

A) \(x = {7 \over 10}- {8 \over10} \)

\(x = {-1 \over 10}\)

B)\({2 \over3}x = 2{5 \over 6}-{3 \over4}\)

\({2 \over3}x = {25 \over 12}\)

\(x = {25 \over 12}/{2 \over3} \)

\(x = {25\over 8}\)

2/ Tính tổng:

\( = {8 \over 9}\)

\(\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)

\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\Rightarrow x=\frac{20}{9}-\frac{2}{9}=\frac{18}{9}=2\)

Vậy x = 2.

k cho mk nha

\(2\frac{2}{9}-x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(2\frac{2}{9}-x=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{3}{9}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{2}{9}\)

\(x=2\frac{2}{9}-\frac{2}{9}\)

\(x=2\)

Vậy x = 2

\(2\frac{2}{9}-x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(2\frac{2}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{9}\)

\(2\frac{2}{9}-x=\frac{2}{9}\)

\(x=2\frac{2}{9}-\frac{2}{9}\)

\(x=2\)

Vậy x = 2