Question

Tìm x

A=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+...+\(\frac{3}{x.\left(x+3\right)}\)=\(\frac{100}{101}\)

Giúp mik vs nha các bạn

Answer 1

\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}\)

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

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

\(=\frac{x+2}{x+3}=\frac{100}{101}\)

\(\Rightarrow x=98\)

Answer 2

\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)

\(A=1-\frac{1}{x+3}=\frac{100}{101}\)

\(\frac{1}{x+3}=1-\frac{100}{101}=\frac{1}{101}\)

=> x + 3 = 101

=> x = 101 - 3

=> x = 98

Vậy x = 98

Ủng hộ mk nha ^_-