Question

[Nguyễn Trung Hiếu]

tính tổng sau bằng cách hợp lý
\(B=\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+...+\frac{a}{x^2+19ax+90a^2}+\frac{1}{x+10a}\)

Answer 1

[Nguyễn Nhật Minh]

\(B=\frac{a}{x\left(x+a\right)}+\frac{a}{\left(x+a\right)\left(x+2a\right)}+\frac{a}{\left(x+2a\right)\left(x+3a\right)}+....+\frac{a}{\left(x+9a\right)\left(x+10a\right)}+\frac{1}{x+10a}\)

  \(=\frac{1}{x}-\frac{1}{x+a}+\frac{1}{x+a}-\frac{1}{x+2a}+\frac{1}{x+2a}-\frac{1}{x+3a}+....+\frac{1}{x+9a}-\frac{1}{x+10a}+\frac{1}{x+10a}\)

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

Answer 2

[Hồng Ánh Lê]

Bạn ơi, thay 1=a nhé!

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