Bài 1:
a: \(A=\left(x-3\right)\left(x^2+3x+9\right)+x^2\left(2-x\right)\)
\(=x^3-27+2x^2-x^3\)
\(=2x^2-27\)
b: \(B=x^2-4-\left(x+2\right)\left(x-1\right)\)
=(x+2)(x-2)-(x+2)(x-1)
=(x+2)(x-2-x+1)
=-1(x+2)
=-x-2
Khi x=-1 thì B=-(-1)-2
=1-2
=-1
Bài 2:
a: \(x^3-4x^2=x^2\cdot x-x^2\cdot4=x^2\left(x-4\right)\)
b: \(3x^2+5y-3xy-5x\)
=3x(x-y)-5(x-y)
=(x-y)(3x-5)
c: \(x^2+10x+25-4y^2\)
\(=\left(x+5\right)^2-\left(2y\right)^2\)
=(x+5+2y)(x+5-2y)
Bài 3:
a: 3(x-2)+4=0
=>3x-6+4=0
=>3x-2=0
=>3x=2
=>\(x=\frac23\)
b: \(x^2-6x+9=0\)
=>\(\left(x-3\right)^2=0\)
=>x-3=0
=>x=3
c: \(x^2-3x+2=0\)
=>\(x^2-x-2x+2=0\)
=>(x-1)(x-2)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)
