a) \(3x^2-3xy-5x+5y\)

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

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

\(=x\left(x-1\right)^2\)

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

A = 5x(x - y) - y(5x - y)

A = 5x² - 5xy - 5xy + y²

A = 5x² - 10xy + y² (1)

Thay x = -1; y = 3 vào (1), ta có:

5.(-1)² - 10.(-1).3 + 3² = 44

B = 4y(x² - 3xy + 3y²) - 2xy(2x - 6y - 3)

B = 4x²y - 12x² + 12y³ - 4x²y + 12xy² + 6xy

B = 12y³ + 6xy (1)

Thay x = 5; y = -1 vào (1), ta có:

12.(-1)³ + 6.5.(-1) = -42

C = 5x²(x - y²) + 3x(xy² - y) - 5x³ 

C = 5x³ - 5x²y² + 3x²y² - 3xy - 5x³ 

C = -2x²y² - 3xy (1)

Thay x = -2; y = -5 vào (1), ta có:

-2.(-2)².(-5)² - 3.(-2).(-5) = -230

D = 6x²(y² - xy + 2x²y) - 3xy(2xy - x² + 4x³)

D = 6x²y² - 6x³y + 12x⁴y - 6x²y² + 3x³y - 12x⁴y

D = -3x³y (1)

Thay x = 11; y = -1 vào (1), ta có:

-3.11³.(-1) = 3993

a. \(3xy+x+15y+15=x\left(3y+1\right)+15\left(y+1\right)=\left(x+15\right)\left(3y+1\right)\)

b.\(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3+x+y\right)\left(3-x-y\right)\)

c.\(x^3-5x^2+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x^2+1\right)\left(x-5\right)\)

d.\(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y+1\right)\left(x-y-1\right)\)

\(\left(5x-1\right)=\left(1-5x\right)^2\)

\(\left(5x-1\right)=\left(5x-1\right)^2\)

\(\left(5x-1\right)\left(1-5x+1\right)=0\)

\(\left(5x-1\right)\left(2-5x\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{2}{5}\end{array}\right.\)

\(9-\left(x-2\right)^2=0\)

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

\(\left(1+x\right)\left(5-x\right)=0\)

\(\left[\begin{array}{nghiempt}x=-1\\x=5\end{array}\right.\)

Bài 4:

1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)

=>\(x^3-1-x^3-6x=11\)

=>-6x-1=11

=>-6x=11+1=12

=>\(x=\dfrac{12}{-6}=-2\)

2: \(16x^2-\left(3x-4\right)^2=0\)

=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)

=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)

=>(x+4)(7x-4)=0

=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)

3: \(x^3-x^2-3x+3=0\)

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

=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-3\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))

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

=>\(x^2+4x+4=x^2-1\)

=>4x+4=-1

=>4x=-5

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)

=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)

=>3x+1=0

=>3x=-1

=>\(x=-\dfrac{1}{3}\left(nhận\right)\)

6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)

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

=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-x-3}{x}=1\)

=>-x-3=x

=>-2x=3

=>\(x=-\dfrac{3}{2}\left(nhận\right)\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

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

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

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

Bài 4: