(x² -y² +6y -9) : (x-y+3)

= [x²-(y²-6y+9)] : (x-y+3)

=[x²-(y-3)² ] : (x-y+3)

=[(x-y+3)(x+y-3)] :(x-y+3)

=x+y-3

\(\left(x^2-y^2+6xy-9\right):\left(x-y+3\right)=\left[\left(x-y\right)^2-9\right]:\left(x-y+3\right)\)

\(=\left(x-y+3\right)\left(x-y-3\right):\left(x-y+3\right)=x-y-3\)

\(x^3-7x-6=0\)

\(x^3-3x^2+3x^2+2x-9x-6=0\)

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

\(\left(x+3\right).\left(x^2+3x+2\right)=0\Rightarrow\left(x-3\right).\left(x^2+3x+x+2\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x+1\right).\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\text{hoặc }x=-2\)

b)B=27y^3-27y^2x+9yx^2-x^3 
= 27 . (1/3x)^3 - 27.(1/3x)².x + 9.1/3.x.x^2 - x^3 
= x^3 - 3x^3 + 3x^3 - x^3 
= 0

d) D=50y^2+x(x-2y)+14y(x-y) 

=50y^2 +x^2 -2xy +14xy -14y^2 

=36y^2 +x^2 +12xy 

=(6y + x)^2 

=81 

Bài 1:

a: =>x^3-x-6x-6=0

=>x(x-1)(x+1)-6(x+1)=0

=>(x+1)(x-3)(x+2)=0

hay \(x\in\left\{-1;3;-2\right\}\)

b: \(\Leftrightarrow x^2-6x+9+y^2+6y+9=0\)

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

=>x=3 và y=-3

(x² - y² + 6y - 9) : (x - y + 3)

= [x² - (y² - 6y + 9)] : (x - y + 3)

= [x² - (y - 3)²] : (x - y + 3)

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

= x + y - 3

\(\left(x^2-y^2+6y-9\right):\left(x-y+3\right)\)

\(=\left[x^2-\left(y^2-6y+9\right)\right]:\left(x-y+3\right)\)

\(=\left[x^2-\left(y-3\right)^2\right]:\left(x-y+3\right)\)

\(=\left(x+y-3\right)\left(x-y+3\right):\left(x-y+3\right)\)

=\(x+y-3\)

ban chu y nhe 
x^2-y^2+6x+9 = (x^2 +6x+9)-y^2(sử dụng hđt a^2+2ab+b^2=(a+b)^2 ) 
=(x+3)^2-y^2 (sử dụng hđt a^2 - b^2= (a+b)(a-b) 
=(x+3+y)(x+3-y) 
=>(x^2 -y^2+6x+9)/(x+y+3)=x-y+3 
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