a) B = \(\frac{\left(a+3\right)^2}{2a^2+6a}\). \(\left(1-\frac{6a-18}{a^2-9}\right)\)
= \(\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\). \(\left(1-\frac{6\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)\)
= \(\frac{a+3}{2a}\). \(\left(1-\frac{6}{a+3}\right)\)
= \(\frac{a+3}{2a}\). \(\frac{a+3-6}{a+3}\)
= \(\frac{a+3}{2a}\). \(\frac{a-3}{a+3}\)
= \(\frac{a-3}{2a}\)
b) B = \(\frac{a-3}{2a}\)= 1
\(\Leftrightarrow\)\(a-3=2a\)
\(\Leftrightarrow\)\(a=-3\)
Vậy khi B = 1 thì a = -3