Theo t/c dãy TSBN:
\(\frac{a-1}{2}=\frac{b-2}{3}=\frac{a-1+b-2}{2+3}=\frac{a+b-3}{5}=\frac{28-3}{5}=\frac{25}{5}=5\)
=> \(\frac{a-1}{2}=5\Rightarrow a-1=5.2=10\Rightarrow a=10+1=11\)
=> \(\frac{b-2}{3}=5\Rightarrow b-2=5.3=15\Rightarrow b=15+2=17\)
Vậy a = 11; b = 17.
Theo tính chất dãy tỉ số bằng nhau ta có: \(\frac{a-1}{2}=\frac{b-2}{3}=\frac{\left(a-1\right)+\left(b-2\right)}{2+3}=\frac{a+b-3}{5}=\frac{28-3}{5}=\frac{25}{5}=5\)
\(\frac{a-1}{2}=5\Leftrightarrow a-1=10\Leftrightarrow a=11\)
\(\frac{b-2}{3}=5\Leftrightarrow b-2=15\Leftrightarrow b=17\)
Vậy a = 11 ; b = 17
