a ) x³ - 9x=0

<=> x (x² - 3 )= 0

<=> x(x+3)(x-3)

<=> x=0

     hoặc x=0-3=-3

      hoặc x=0+3=3

x(x+5)+(x+1)^2

=x(x+5)+x^2+2x+1

=x^2+5x+x^2+2x+1

=2x^2+7x+1

a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)

\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)

\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)

\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)

\(=\left(x-3\right)\left(2x^2-11x\right)\)

\(=x\left(x-3\right)\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)

\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)

\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)

\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)

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

\(=\left(x-2y+5\right)\left(x+2y+1\right)\)

a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)

\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)

b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)

 \(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)

\(a\left(a^2-bc\right)+b\left(b^2-ca\right)+c\left(c^2-ab\right)=0\)

\(\Rightarrow a^3-abc+b^3-abc+c^3-abc=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\) 

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

Mà \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow}a=b=c\)

Vậy \(P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)