\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)

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

=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)

=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)

=\(\frac{x}{x-3}\)

= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]

 \(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)

= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)

= \(\frac{x+3}{x-3}\)

k mik nhé. Plssss~

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)

x= {7;9} **** mk nha 

(9x-63).(27-3x)=0(x> hoặc =7)

=> 9x-63=0 hoặc 27-3x=0

tính lần lượt ra loại 2-x vì x> hoặc =7 =>x=7

ĐKXĐ: \(x\ne\pm3\)

\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)

Ý 2 mình k hiểu ý bạn lắm

\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)

a) 9x (3x-y) +3y (y-3x)

= 9x(3x-y)-3y(3x-y)

=(3x-y)(9x-3y)

= (3x-y)3(3x-y)

= 3(3x-y)²

b)x³ - 3x² - 9x + 27

=x³+3x²-6x²-18x+9x+27

= (x³+3x²)-(6x²+18x)+(9x+27)

= x²(x+3)-6x(x+3)+9(x+3)

= (x+3)(x²-6x+9)

=(x+3)(x-3)²

a) \(9x\left(3x-y\right)+3y\left(y-3x\right)=9x\left(3x-y\right)-3y\left(3x-y\right)\)

=\(\left(3x-y\right)\left(9x-3y\right)\)

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

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

b) \(x^3-3x^2-9x+27=\left(x^3-3x^2\right)-\left(9x-27\right)\)

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

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

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

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

\(\text{a) }9x\left(3x-y\right)+3y\left(y-3x\right)\\ \\ =9x\left(3x-y\right)-3y\left(3x-y\right)\\ \\ =3\left(3x-y\right)\left(3x-y\right)\\ \\ =3\left(3x-y\right)^2\)

\(\text{b) }x^3-3x^2-9x+27\\=\left(x^3-3x^2\right)-\left(9x-27\right) \\=x^2\left(x-3\right)-9\left(x-3\right)\\ \\=\left(x^2-9\right)\left(x-3\right)\\ =\left(x+3\right)\left(x-3\right)\left(x-3\right)\\ \\ =\left(x+3\right)\left(x-3\right)^2\)

\(x^3-3x^2-9x+27\)

\(x^3+27-3x^2-9x\)

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

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

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