Question

Rút gọn:
\(A=\sqrt{\left(a-3\right)^2}-3a\) với a < 3

\(B=4a+3-\sqrt{\left(2a-1\right)^2}\) với a > 1/2

\(C=\dfrac{4}{a^2-4}\sqrt{\left(a-2\right)^2}\) với a < 2

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{a^2+6a+9}{16}}\) với a < -3

Answer 1

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

Answer 2

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a