Giải:

\(A=1\times2+2\times3+3\times4+...+99\times100\)

\(3A=3\times\left(1\times2+2\times3+3\times4+...+99\times100\right)\)

\(3A=1\times2\times3+2\times3\times3+3\times4\times3+...+99\times100\times3\)

\(3A=1\times2\times3+2\times3\times\left(4-1\right)+3\times4\times\left(5-2\right)+...+99\times100\times\left(101-98\right)\)

\(3A=1\times2\times3+2\times3\times4-1\times2\times3+...+99\times100\times101-98\times99\times100\)

\(\rightarrow3A=99\times100\times101=999900\)

\(\Rightarrow A=999900\div3=333300\)

Vậy \(A=333300\)

Thay  999990=a vào biểu thức A ta được

\(A=\left(a+4\right)\left(a+9\right)\left(a+2\right)-\left(a+6\right)\left(a+1\right)\left(a+8\right)\)

\(=\left(a^2+13a+36\right)\left(a+2\right)-\left(a^2+7a+6\right)\left(a+8\right)\)

\(=a^3+2a^2+13a^2+26a+36a+72-a^3-8a^2-7a^2-56a-6a-48\)

\(=24\)

Thay b=44440 vào B ta được

\(B=\left(b+3\right)\left(b+8\right)\left(b+1\right)-\left(b+5\right)b\left(b+7\right)\)

\(=\left(b^2+11b+24\right)\left(b+1\right)-\left(b^2+5b\right)\left(b+7\right)\)

\(=b^3+b^2+11b^2+11b+24b+24-b^3-7b^2-5b^2-35b\)

\(=24\)

Vậy A=B (=24)