Ta có  \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

           \(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

                 \(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)

                 \(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

                 \(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

                 \(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)

                  \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

                   \(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

                     \(\frac{1}{x+1}=\frac{1}{18}\left(1\right)\)

                   Có \(\left(1\right)\Leftrightarrow\left(x+1\right).1=1.18\) 

                                       \(\Rightarrow x+1=18\)

                                       \(\Rightarrow x=18-1\)

                                       \(\Rightarrow x=17\)

\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{9}\)

\(\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}+...+\frac{\left(X+1\right)-X}{Xx\left(X+1\right)}=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{X+1}=\frac{1}{9}\)

\(\frac{1}{X+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

X+1=18

X=17

Ta có : 

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

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

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

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

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

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

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

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

\(\Leftrightarrow\)\(x-2=\frac{8}{1}.\frac{1}{1}\)

\(\Leftrightarrow\)\(x-2=8\)

\(\Leftrightarrow\)\(x=8+2\)

\(\Leftrightarrow\)\(x=10\)

Vậy \(x=10\)

Chúc bạn học tốt ~ 

(x+x+x+x+x+x)-(2/12+2/20+2/30+2/42+2/56+2/72)=16/9
6x-4/9=16/9
6x=20/9=>x=20/7

Trả lời

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

\(x-2\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{21}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

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

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

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

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

\(\)\(\Rightarrow2\left(x-2\right)=16\)

\(\Rightarrow x-2=16:2\)

\(\Rightarrow x-2=8\)

\(\Rightarrow x=8+2\)

\(\Rightarrow x=10\)

Vậy x=10

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

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

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

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

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

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

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

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

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

\(x-2=8\)

\(x=8+2\)

\(x=10\)

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
 

=> 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

2.[1/42+1/56+1/72+...+1/x.(x+1)]=2/9

1/6.7+1/7.8+1/8.9+....+1/x.(x+1)=1/9

1/6-1/7+1/7-1/8+1/8-1/9+.....+1/x-1/x+1=1/9

1/6-1/x+1=1/9

1:(x+1)=1/6-1/9

x+1=1:(1/18)

x+1=18

x=18-1

x=17

Vậy x=17

Chúc em học tốt

Ủng hộ anh nha^^

2/42 + 2/56 + 2/72 + ... + 2/x.(x+1) = 2/9

2.[1/42 + 1/56 + 1/72 + ... + 1/x.(x+1)] = 2/9

1/6.7 + 1/7.8 + 1/8.9 + ... + 1/x.(x+1) = 2/9 : 2

1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x+1 = 2/9 . 1/2

1/6 - 1/x+1 = 1/9

1/x+1 = 1/6 - 1/9

1/x+1 = 6/36 - 4/36

1/x+1 = 2/36 = 1/18

=> x+1=18

=> x=18-1

=> x=17

Vậy x=17

Đặt VT là A ta có:

\(A=2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

\(=2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)

Thay A vào ta có:\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

\(\Rightarrow x+1=18\)

\(\Rightarrow x=17\)

ta xét VT=\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{x\left(x+1\right)}\right)\)

                =\(2\left(\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

                =\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)= 2*1/9

=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

<=> \(\frac{1}{x+1}=\frac{1}{18}\)

<=> x+1=18

=> x=17

tớ làm khác nhưng kết quả thì giống