Question

Answer 1

Bài 1:

a: \(a^2x^2-16y^2=\left(a\cdot x\right)^2-\left(4y\right)^2=\left(ax-4y\right)\left(ax+4y\right)\)

b: \(9x^2-16\left(x+1\right)^2\)

\(=\left(3x\right)^2-\left(4x+4\right)^2\)

=(3x-4x-4)(3x+4x+4)

=(-x-4)(7x+4)

c: \(x^2-3=\left(x-\sqrt3\right)\left(x+\sqrt3\right)\)

d: \(3x^4y-12x^3y+12x^2y\)

\(=3x^2y\left(x^2-4x+4\right)=3x^2y\left(x-2\right)^2\)

Bài 2:

a: \(3x+18-9x\left(x+6\right)=0\)

=>3(x+6)-9x(x+6)=0

=>(x+6)(3-9x)=0

=>\(\left[\begin{array}{l}x+6=0\\ 3-9x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ 9x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=\frac13\end{array}\right.\)

b: \(2x\left(x-5\right)-x\left(2x+3\right)=26\)

=>\(2x^2-10x-2x^2-3x=26\)

=>-13x=26

=>x=-2

c: \(\left(x-3\right)^2-x\left(x+4\right)=-11\)

=>\(x^2-6x+9-x^2-4x=-11\)

=>-10x=-20

=>x=2

d: \(3x^2-6x=0\)

=>3x(x-2)=0

=>x(x-2)=0

=>x=0 hoặc x=2