a)
\(A=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|\)
TH1 x-3>0 => x>3
\(A=6-2x-x+3=9-3x\)
TH2 x-3<0 => x<3
\(A=6-2x+x-3=3-x\)
b)thay x = 5 ta có
TH1 A = 9 -3 . 5 = -6
TH2 A =3 - 5 = -2
c)
TH1
9-3x=0
<=> 3(3-x) = 0
<=> 3x=0
x=0
TH2
3-x=0
-x=-3
x=3
\(a,A=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|\\ b,x=5\Leftrightarrow A=6-10-7=-11\\ c,A=0\Leftrightarrow\left|x-3\right|=6-2x\\ \Leftrightarrow\left[{}\begin{matrix}x-3=6-2x\left(x\ge3\right)\\x-3=2x-6\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\)
a: \(A=6-2x-\sqrt{x^2-6x+9}\)
\(=6-2x-\left|x-3\right|\)
\(=\left[{}\begin{matrix}6-2x-x+3=-3x+9\left(x\ge3\right)\\6-2x+x-3=-x+3\left(x< 3\right)\end{matrix}\right.\)
b: Vì x=5>3 nên A=-3x5+9=-15+9=-6
