\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)

a: \(y^3+2xy^2+y^2-4x^2\)

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

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

\(\frac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\)

\(=\frac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{\left(2x+y\right)\left(y^2+y-2x\right)}=\frac{4x^2-2xy+y^2}{y^2+y-2x}\)

b: \(\frac{x^2-2x-8}{2x^2+9x+10}\)

\(=\frac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

\(=\frac{\left(x-4\right)\cdot\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}=\frac{x-4}{2x+5}\)

c: \(\frac{6x-x^2-5}{5x^6-x^7}\)

\(=\frac{x^2-6x+5}{x^7-5x^6}\)

\(=\frac{\left(x-5\right)\left(x-1\right)}{x^6\cdot\left(x-5\right)}=\frac{x-1}{x^6}\)

d: \(\frac{x^3+64}{2x^3-8x^2+32x}=\frac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}=\frac{x+4}{2x}\)

e: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

\(=\frac{x+y}{\left(x-y\right)\left(x+y\right)}=\frac{1}{x-y}\)

6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

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

8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)

10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)

6) 9x³y² + 3x²y² = 3x²y²( 3x + 1 )

7) x³ + 2x² + 3x = x( x² + 2x + 3 )

8) 6x²y + 4xy² + 2xy = 2xy( 3x + 2y + 1 )

9) 5x²( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )

10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )

mình ghi bị nhầm bài rồi

vậy bạn ghi lại bài đúng đi

do hơi bận nên mk ghi đáp án nha, ko hiểu đâu ib mk 

a)  \(3xy^2-2xy+12x=x\left(3y^2-2y+12\right)\)

b)  \(x^3-10x^2+25x-16xy^2=x\left(x-4y-5\right)\left(x+4y-5\right)\)

c)  \(5y^3-10xy^2+5x^2y-20y=5y\left(y-x-2\right)\left(y-x+2\right)\)

d)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)\left(x+y-z\right)\) 

e)  \(9x^2+y^2+6xy=\left(3x+y\right)^2\)

f)  \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

g)  \(125x^3-75x^2+15x-1=\left(5x-1\right)^3\)

h)  \(x^2-xz-9y^2+3yz=\left(x-3y\right)\left(x+3y-z\right)\)

a: \(9x^3y^2+3x^2y^2\)

\(=3x^2y^2\cdot3x+3x^2y^2=3x^2y^2\left(3x+1\right)\)

b: \(x^3+2x^2+3x=x\cdot x^2+x\cdot2x+x\cdot3=x\left(x^2+2x+3\right)\)

c: \(6x^2y+4xy^2+2xy\)

\(=2xy\cdot3x+2xy\cdot2y+2xy\)

=2xy(3x+2y+1)

d: \(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

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

a) Ta có: \(5y^3-10xy^2+5yx^2-20y\)

\(=5y\left(y^2-2xy+x^2-4y\right)\)

b) Ta có: \(x^2+2xy+y^2-xz-yz\)

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

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

c) Ta có: \(9x^2+y^2+6xy\)

\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)

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

d) Ta có: \(8-12x+6x^2-x^3\)

\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)

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

e) Ta có: \(125x^3-75x^2+15x-1\)

\(=\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot1+3\cdot5x\cdot1^2-1^3\)

\(=\left(5x-1\right)^3\)

Đề bài là gì sao không ghi rõ?? 

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^