Bài 1:
a) Ta có: \(A=\left(x+3\right)^2+2\left(x+3\right)\left(x-3\right)-3x\left(x-3\right)\)
\(=x^2+6x+9+2\left(x^2-9\right)-3x^2+9x\)
\(=-2x^2+15x+9+2x^2-18\)
\(=15x-9\)
b) Ta có: \(\dfrac{2x}{3x-3y}-\dfrac{x^2}{x-y}\)
\(=\dfrac{2x}{3\left(x-y\right)}-\dfrac{3x^2}{3\left(x-y\right)}\)
\(=\dfrac{2x-3x^2}{3\left(x-y\right)}\)
c) Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=0\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)