Question

Tìm x : \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\cdot\left(x+1\right)}=\frac{2}{9}\) ( x thuộc N*)

Answer 1

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^^

Answer 2

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

Answer 3

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

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

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

\(\Rightarrow x+1=18\)

\(\Rightarrow x=17\)