Question

1/3x7+1/7x11+...+1/Xx[X+4]=5/63

 

tìm X

Answer 1

Ta có : \(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x.\left(x+4\right)}=\frac{5}{63}\)

=> \(\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{5}{63}\)

=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{20}{63}\)

=> \(\frac{1}{3}-\frac{1}{x+1}=\frac{20}{63}\)

=> \(\frac{1}{x+1}=\frac{1}{63}\)

=> x + 1 = 63

=> x = 62

Vậy x = 62

Answer 2

Sửa lại bài làm của XYZ một chút:

=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+..+\)\(\frac{1}{x}-\frac{1}{x+4}\)

=> \(\frac{1}{3}-\frac{1}{x+4}\)= \(\frac{5}{63}\div\frac{1}{4}=\frac{20}{63}\)f

Answer 3

Làm tiếp:

=> \(\frac{1}{x+4}=\frac{1}{3}-\frac{20}{63}\)

=> \(\frac{1}{x+4}=\frac{1}{63}\)

=> x + 4 = 63

=> x = 63 - 4

=> x = 59