Bài `2`

`a, 2x^3 -6x^2`

`= 2x^2(x- 3)`

`b, 9x^2 -1/16y^2`

`= (3x)^2 -(1/4y)^2`

`=(3x-1/4y)(3x+1/4y)`

`c, x^2y +5xy^2 +6y^3`

`= y ( x^2 + 5xy+6y^2)`

`d, 3x^3 +6x^2y`

`= 3x^2 (x+ 2y)`

`e, -9+6x-x^2`

`= -(x^2-6x+9)`

`=-(x-3)^2`

`f, 2x^2 +3xy -5y^2`

`= x(2x+ 3y- 5y)`

`g, x^2 +12x+36`

`= x^2 + 2.x.6+6^2`

`=(x+6)^2`

`h, x^2 +1-y^2+2x`

`=(x^2 +2x+1)-y^2`

`=(x+1)^2-y^2`

`=(x+1-y)(x+1+y)`

`i, 4x^2 +12x+9`

`=(2x)^2 + 2.2x.3+3^2`

`=(2x+3)^2`

`j, 25-x^2 +2xy-y^2`

`= -(x^2 +2xy -y^2 +25)`

`= -x^2 -2xy +y^2 -25`

`=-(x-y)^2 -25`

`=-(x-y-5)(x-y+5)`

`k,x^2-25`

`=x^2-5^2`

`=(x-5)(x+5)`

`l, 3x^3y-12x^2y^2+12xy^3`

`= 3xy( x^2 - 4xy + 4y^2)`

`= 2xy( x-2y)^2`

`5x^5 =x^3`

`=> 5x^5-x^3=0`

`=> x^3 ( 5x^2-1)=0`

\(\Rightarrow\left[{}\begin{matrix}x^3=0\\5x^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)

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\(\left(2x-3\right)^3=64\\ \Rightarrow\left(2x-3\right)^3=4^3\\ \Rightarrow2x-3=4\\ \Rightarrow2x=4+3\\ \Rightarrow2x=7\\ \Rightarrow x=\dfrac{7}{2}\left(ktm\right)\)

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`(5x-1)^2 =2023^0 . 7 +3^2`

`=> (5x-1)^2 =1. 7 +9`

`=> (5x-1)^2 =7+9`

`=> (5x-1)^2 =16`

`=> (5x-1)^2 = (+- 4)^2`

\(\Rightarrow\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x=5\\5x=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)

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KTM ( Không thỏa mãn ) vì `x` là số tự nhiên.