\(\frac{A}{11}+\frac{B}{3}=\frac{31}{33}\)
=> \(\frac{3A}{33}+\frac{11B}{33}=\frac{31}{33}\)
=> 3A + 11B = 31
=> 3A + 11B = 9 + 22 = 3 x 3 + 11 x 2
=> A = 3; B = 2
Theo đề bài ra ta có:
\(\frac{A}{11}+\frac{B}{3}=\frac{31}{33}\Rightarrow\frac{A\times3+B\times11}{11\times3}=\frac{31}{33}Mà31=3\times3+2\times11\Rightarrow A=3;B=2\)