b: \(=x^2+6x+9-x^2+4=6x+13\)

\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)

a)4x²+8x                     b)x²-9

=4x(x+2)                      =x²-3²

                                     =(x-3)(x+3)

c)2x³-3x²+2x-3

=2x³+2x-(3x²+3)

=2x(x²+1)-3(x²+1)

=(2x-3)(x²+1)

\(x^2\left(2x+3\right)-2x^3=3\\ \Leftrightarrow2x^3+3x^2-2x^3=3\\ \Leftrightarrow3x^2=3\\ \Leftrightarrow x^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(a,6\left(x-2\right)=8\left(3x+1\right)\\ \Leftrightarrow6x-12=24x+8\\ \Leftrightarrow18x+20=0\\ \Leftrightarrow x=-\dfrac{10}{9}\\ b,2x-\left(3-7x\right)=5\left(x+3\right)\\ \Leftrightarrow2x-3+7x=5x+15\\ \Leftrightarrow9x-3-5x-15=0\\ \Leftrightarrow4x-18=0\\ \Leftrightarrow x=\dfrac{9}{2}\\ c,\left(x-1\right)^2=\left(x+3\right)\left(x+2\right)\\ \Leftrightarrow x^2-2x+1=x^2+5x+6\\ \Leftrightarrow7x+5=0\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d,\left(3x-9\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(e,x^2-3x+2=0\\ \Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\\ \Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ f,x^2-4x+4=0\\ \Leftrightarrow x^2-2.2+2^2=0\\ \Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ x=2\)

a, \(6x-12=24x+8\Leftrightarrow18x=-20\Leftrightarrow x=-\dfrac{20}{18}=-\dfrac{10}{9}\)

b, \(2x-3+7x=5x+15\Leftrightarrow4x=18\Leftrightarrow x=\dfrac{9}{2}\)

c, \(x^2-2x+1=x^2+5x+6\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)

d, \(\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

e, \(x^2-3x+2=0\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow x=1;x=2\)

f, \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

\(\Leftrightarrow x-2=0\)

hay x=2

\(\text{x^2-4x+4=0}\\ \left(x-2\right)^2=0\\ x-2=0\\ x=0+2\\ x=2\)

2

c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)

hay x=1

b) 4x(2-x)+(2x+1)^2=2
    8x-4x^2+4x^2+4x+1-2=0
    (8x+4x)+(-4x^2+4x^2)+(1-2)=0
         12x + 0 -1 =0
                 12x=1
                     x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
     x^3-9x^2+27x-x^3+9x^2=0
     (x^3-x^3)+(-9x^2+9x^2)+27x=0
           0 + 0 + 27x=0
                     x= 0
Vậy x=0
    

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{3;-3\right\}\)

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

\(TH1\left(x-9\right)=0\Leftrightarrow x=9\)

\(TH2\left(x+9\right)=0\Leftrightarrow x=-9\)