- Khái niệm và cách chứng minh định lí Menelaus:

https://vi.wikipedia.org/wiki/%C4%90%E1%BB%8Bnh_l%C3%BD_Menelaus

Bài 12:

1) A=\(\dfrac{636363.37-373737.63}{1+2+3+...+2006}\)=\(\dfrac{10101.63.37-10101.37.63}{1+2+3+...+2006}\)=\(\dfrac{0}{1+2+3+...+2006}\)=0

Bài 11:

a. A=\(\dfrac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)=\(\dfrac{1.5.6+2^3.\left(1.5.6\right)+4^3.\left(1.5.6\right)+9^3.\left(1.5.6\right)}{1.3.5+2^3.\left(1.3.5\right)+4^3.\left(1.3.5\right)+9^3.\left(1.3.5\right)}\)=\(\dfrac{1.5.6\left(1+2^3+4^3+9^3\right)}{1.3.5\left(1+2^3+4^3+9^3\right)}=2\)

b. \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\)

=\(k\left(k+1\right)\left(k+2-k+1\right)\)=\(3k\left(k+1\right)\)

S=\(1.2+2.3+3.4+...+n\left(n+1\right)\)

3S=\(1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3\)

3S=\(1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right).\left(n+2-n+1\right)\)

3S=\(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right).\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)S=\(\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)