\(A=\sqrt{1296\left(1-a\right)^2}=36\left|1-a\right|=36\left(a-1\right)\\ B=\dfrac{1}{a-b}\cdot a^2\left|a-b\right|=\dfrac{a^2\left(a-b\right)}{a-b}=a^2\\ C=\sqrt{225a^2}-3a=\left|15a\right|-3a=15a-3a=12a\\ D=9-6a+a^2-\sqrt{36a^2}=9-6a+a^2-\left|6a\right|\\ =\left[{}\begin{matrix}9-6a+a^2-6a\left(a\ge0\right)\\9-6a+a^2+6a\left(a< 0\right)\end{matrix}\right.=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\)
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