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)

4x^2 + 8x

= 4x . x + 4x . 2

= 4x ( x + 2 )

\(4x^{^{ }2}+8x\) 

\(=4x.x+4x.2\)

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

\(a,\left(x-2\right)^2=x^2-2.x.2+2^2=x^2-4x+4\\ b,\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\\ c,x^2-5x^2=-4x^2\)

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

\(2x^2+7x-9=0\Leftrightarrow\left(x-1\right)\left(x+\dfrac{9}{2}\right)=0\)

\(x\in\left\{1;\dfrac{-9}{2}\right\}\)

<=> 2x² -2x +9x -9 =0

<=> 2x(x-1) + 9 (x-1) = 0

<=> (2x+9)(x-1) = 0

<=> 2x + 9 =0 hoặc x-1 = 0

<=> x= -9/2 hoặc x= 1

\(\Leftrightarrow x-2=0\)

hay x=2

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

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