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

2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)

a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)

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

\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)

a: \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)

c: \(x^2-3x+3y-y^2\)

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

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

b: \(x^2-6x-7=\left(x-7\right)\left(x+1\right)\)

a) \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)

b) \(x^2-6x-7=x\left(x-7\right)+\left(x-7\right)=\left(x-7\right)\left(x+1\right)\)

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

d) \(x^3-xy+2y-8=\left(x-2\right)\left(x^2+2x+4\right)-y\left(x-2\right)\)

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

a)x(2x - y) - y(2x - y)

=  (x - y)(2x - y)

c)x² - 3x + 3y - y² 

= x² - y² - 3x + 3y

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

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

b)x² - 6x - 7

= x² + x - 7x - 7

= x(x + 1) - 7(x + 1)

= (x - 7)(x + 1)

d)x³ - xy + 2y - 8

= x³ - 8 - xy + 2y

= x³ - 2³ - y(x - 2)

= (x - 2)(x² + 2x + 4) - y (x - 2)

= (x - 2 - y)(x² + 2x + 4)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

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Chọn B.

+ Ta có các số 5x - y; 2x + 3y; x + 2y lập thành CSC nên suy ra

2( 2x + 3y) = 5x – y + x+ 2y  hay 2x = 5y  (1)

Các số (y + 1)², xy + 1, (x – 1)² lập thành CSN suy ra (xy + 1)² = (y + 1)²(x – 1)² ⇔ (4 + 2y – 2x)(4xy + 2x – 2y) = 0 (2)

Thay (1) vào (2) ta được: (4 + 2y – 5y)(10y² + 5y – 2y) = 0

⇔ y(4 – 3y)(10y + 3) = 0 ⇔ y = 0, y = 4/3, y = -3/10.

Vậy 

a) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{1+x+1}\) \(=\dfrac{x^2.\left(x-1\right)\left(x+2\right)}{\left(x+1\right).\left(x-1\right)\left(x+2\right)}+\dfrac{2x.\left(x+2\right)}{\left(x-1\right).\left(x+1\right).\left(x+2\right)}+\dfrac{\left(x-1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)

\(=\dfrac{x^2.\left(x-1\right).\left(x+2\right)+2x.\left(x+2\right)+\left(x-1\right)\left(x+1\right)}{\left(x+1\right).\left(x-1\right).\left(x+2\right)}\)

\(=\dfrac{x^4+x^3-2x^2+2x^2+4x+x^2-1}{\left(x-1\right)\left(x+1\right).\left(x+2\right)}\)

\(=\dfrac{x^4+x^3+x^2+4x-1}{\left(x^2-1\right).\left(x+2\right)}\)

\(=\dfrac{x^4+x^3+x^2+4x-1}{x^3+2x^2-x-2}\)

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

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

b, \(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x^2}{x^2-1}=\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\)

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

Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$