Question

Answer 1

1A:

a: \(3x-3y+x^2-y^2\)

=3(x-y)+(x-y)(x+y)

=(x-y)(x+y+3)

b: \(x^6-x^4+2x^3+2x^2\)

\(=x^2\left(x^4-x^2+2x+2\right)\)

\(=x^2\left\lbrack x^2\left(x^2-1\right)+2\left(x+1\right)\right\rbrack\)

=\(x^2\left(x+1\right)\left(x^3-x+2\right)\)

b: \(x^2-4x^2y^2+y^2+2xy\)

\(=\left(x+y\right)^2-\left(2xy\right)^2\)

=(x+y-2xy)(x+y+2xy)

d: \(x^3-3x^2+3x-1-y^3\)

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left\lbrack\left(x-1\right)^2+y\cdot\left(x-1\right)+y^2\right\rbrack\)

\(=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)

1B:

a: \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)

\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left\lbrack\left(x-3\right)^2-1\right\rbrack\)

=(x-1)(x+1)(x-3-1)(x-3+1)

=(x-1)(x+1)(x-4)(x-2)

b: \(x^3-2x^2+4x-8\)

\(=x^2\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+4\right)\)

c: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-\left(x^3-3x^2y+3xy^2-y^3\right)\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)

d: \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=2\left(x+y+z\right)\left(a^2-2ab+b^2\right)\)

\(=2\left(x+y+z\right)\left(a-b\right)^2\)

2A:

a: \(3x\left(x-1\right)+\left(x-1\right)=0\)

=>(x-1)(3x+1)=0

=>\(\left[\begin{array}{l}x-1=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac13\end{array}\right.\)

b: \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

=>\(\left(x-2\right)\left\lbrack x^2+2x+7+2\left(x+2\right)-5\right\rbrack=0\)

=>(x-2)\(\left(x^2+2x+7+2x+4-5\right)=0\)

=>(x-2)(\(x^2+4x+6\) )=0

=>x-2=0

=>x=2

c: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

=>(x-3)(x+2)=0

=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

d: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

=>(x+3)(\(x^2-3x+9+x-9\) )=0

=>(x+3)(\(x^2-2x\) )=0

=>x(x-2)(x+3)=0

=>x∈{0;2;-3}