Question

Chứng minh

\(\frac{2}{\left(n-1\right)\cdot n\left(n+1\right)}=\frac{1}{n\cdot\left(n-1\right)}-\frac{1}{n\cdot\left(n+1\right)}\)

 

Answer 1

VP:

\(\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)}{\left[n\left(n-1\right)\right]\left[n\left(n+1\right)\right]}-\frac{n\left(n-1\right)}{\left[n\left(n-1\right)\right]\left[n\left(n+1\right)\right]}\)

\(=\frac{n^2+n}{\left(n^2-n\right)\left(n^2+n\right)}-\frac{n^2-n}{\left(n^2-n\right)\left(n^2+n\right)}\)

\(=\frac{\left(n^2+n\right)-\left(n^2-n\right)}{\left(n^4-n^3+n^3-n^2\right)-\left(n^4-n^3+n^3-n^2\right)}\)

\(=\frac{2n}{\left(n^4-n^2\right)-\left(n^4-n^2\right)}\)

\(=\frac{2n}{0}\)

Ủa! Hình như tớ lm sai ở đâu đó.