a) \(6x-6y=6\left(x-y\right)\)

b)\(2xy+3x+6y+xz\)

\(=\left(2xy+xz\right)+\left(6y+3z\right)\)

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

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

c)\(x^2+6x+9-y^2\)

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

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

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

d) \(9x-x^3\)

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

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

e)\(x^2-xy+x-y\)

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

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

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

a, 6x - 6y = 6( x-y )

b, 2xy + 3z + 6y + xz

  = ( 2xy + 6y ) + ( 3z + xz )

  = 2y( x + 3 ) + z ( 3 + x )

  = 2y( 3 + x ) + z ( 3 + x )

  = ( 3 + x ) ( 2y + z )

c, x² + 6x + 9 - y² = ( x² + 6x + 9 ) - y²

                             = ( x + 3 )² - y²

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

d , 9x - x³ = x ( 9 - x² )

                = x ( 3 - x ) ( 3 + x )

e, x² - xy + x - y =( x ² - xy ) + ( x - y )

                          = x ( x - y ) + ( x - y )

                          = ( x - y ) ( x + 1 )

\(a,6x-6y=6\left(x-y\right)\)

\(b.2xy+3z+6y+xz\)

\(2y\left(x+3\right)+z\left(x+3\right)\)

\(\left(2y+z\right)\left(x+3\right)\)

\(c,x^2+6x+9-y^2\)

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

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

\(d,9x-x^3\)

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

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

\(e,x^2-xy+x-y\)

\(x\left(x+1\right)-y\left(x+1\right)\)

\(\left(x+1\right)\left(x-y\right)\)

A=5x^2+6x^2+3y+7y=11x^2+10y

B=7x^3+6x^3+6y+5y+36=13x^3+11y+36

C=-8x^5-x^5+3y^4-10y^4=-9x^5-7y^4

C=x^2-5x^2+y^2-6y^2=-4x^2-5y^2

Ta có \(\frac{1+2y}{18}\)=\(\frac{1+4y}{24}\)

\(\Rightarrow\)(1+2y)24=(1+4y)18

\(\Rightarrow\)24+48y=18+72y

\(\Rightarrow\)24-18=72y-48y

\(\Rightarrow\)6=24y

\(\Rightarrow\)y=\(\frac{6}{24}\)

\(\Rightarrow\)y=\(\frac{1}{4}\)

Thay y=\(\frac{1}{4}\) vào đề ta có:

1 + 2\(\frac{1}{4}\) / 18 = 1 + 4\(\frac{1}{4}\) / 24 = 1 + 6\(\frac{1}{4}\) / 6x

=>\(\frac{1}{12}\)=\(\frac{\frac{5}{2}}{\frac{6}{x}}\)

=>12.\(\frac{5}{2}\)=6x

=>30=6x

=>x=5

Vậy x=5;y=\(\frac{1}{4}\)

ta co : 1+2y/18=1+4y/24

=> 24(1+2y)=18(1+4y)

=>24+48y=18+72y

=>24-18=72y-48y

=>6=24y

=>y=1/4 

thay y thanh 1/4 vao de bai ta co : 

1+1/2/18=1+1/24=(1+3/2)/6x

=>1/12=(5/2)/6x

=>12/(5/2)=6x

=>30=6x/x=5

vay x=5 va y=1/4

Ta có =

(1+2y)24=(1+4y)18

24+48y=18+72y

24-18=72y-48y

6=24y

y=

y=

Thay y= vào đề ta có:

1 + 2 / 18 = 1 + 4 / 24 = 1 + 6 / 6x

=>=

=>12.=6x

=>30=6x

=>x=5

Vậy x=5;y=

\(xy+yz+zx=8xyz\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=8\)

\(\Rightarrow\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}=64\)

Ta có: \(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\)

\(=\left(\dfrac{1}{x}+...+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{y}+...+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}\right)+\left(\dfrac{1}{z}+...+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}\right)\)

(sau dấu chấm là bốn số tương tự).

\(\ge^{Cauchy-Schwarz}\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow64\ge\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow\dfrac{1}{6x+y+z}+\dfrac{1}{6y+z+x}+\dfrac{1}{6z+x+y}\le1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{8}\)

Vậy \(Max\) của biểu thức đã cho là 1.