dienlanhquanly

Thêm mấy cái
$(a+b)(b+c)(c+a)-8abc=a(b-c)^2+b(c-a)^2+c(a-b)^2$

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}-\dfrac{3}{2}=\sum _{sym} \dfrac{(a-b)^2}{2(c+a)(b+c)}$

méo hiểu

Cái này thua rồi bạn oi . hiuhu

Thêm mấy cái nữa nhé:
13+23+33+.....n3=(1+2+3+...+n)213+23+33+.....n3=(1+2+3+...+n)2
1.2+2.3+3.4+4.5+....+n(n+1)=n(n+1)(n+2)31.2+2.3+3.4+4.5+....+n(n+1)=n(n+1)(n+2)3
12+22+32+...+n2=n(n+1)(2n+1)612+22+32+...+n2=n(n+1)(2n+1)6 
1.n+2(n−1)+3(n−2)+.....+n.1=n(n+1)(n+2)61.n+2(n−1)+3(n−2)+.....+n.1=n(n+1)(n+2)6