\(A=x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\\ =x^3-x^2y^2-xy+x^2y^2-x^3\\ =\left(x^3-x^3\right)+\left(-x^2y^2+x^2y^2\right)-xy\\ =-xy\)

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

\(=x^3-x^2y^2-xy+x^2y^2-x^3\)

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

\(=-xy\)

Vậy \(A=-xy\)

#\(Toru\)

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

Bài 1:

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

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

\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)

\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).

a) 3.(x+y).(x-y)+(x+y)^2+(x-y)^2

=3.(x²-y²)+(x²+2xy+y²)+(x²-2xy+y²)

=3x²-3y²+x²+2xy+y²+x²-2xy+y²

=5x²-y²
b) (2x+y)^2 - (y+3x)^2

=[(2x+y)+(y+3x)][(2x+y)-(y+3x)]

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

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

=-5x²-2xy

a) \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

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

\(=5.\left(-2xy^2\right)\)

\(=-10xy^2\)

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

\(=x^3-y^3\)

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

\(=x^3-3x^2.3+3x.3^2-3^3+2^3-3.2^2.x+3.2.x^2-x^3\)

\(=x^3-9x^2+27x-27+8-12x+6x^2-x^3\)

\(=\left(x^3-x^3\right)+\left(-9x^2+6x^2\right)+\left(27x-12x\right)+\left(-27+8\right)\)

\(=-3x^2+15x-19\)

a) (x+3)(x^2-3x+9)-(54+x^3)

= x^3- 3x^2+9x+3x^2-9x+27-54-x63

= -27

b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)

= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]

= [(2x)3^3+ y^3] – [(2x)^3 – y^3]

= (2x)^3 + y^3 – (2x)^3 + y^3

= 2y^3

a)(x+3)(X^2-3x+9)-(54+x^3)

= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)

= 27- 54

= -27

b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)

= \((2x)^3\) + \(y^3\)  - [\((2x)^3\) - \(y^3\) ]

= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)

= \(2y^3\)

a) Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+27-54-x^3\)

=-27

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

A = (\(x-y\)).(\(x^2\) + \(xy\) + y²) + 2y³

A = \(x^3\) - y³ + 2y³

A = \(x^3\) + y³

Thay \(x=\dfrac{2}{3}\); y = \(\dfrac{1}{3}\) vào biểu thức

A = \(x\)³ +  y³ ta có:

A = (\(\dfrac{2}{3}\))³ + (\(\dfrac{1}{3}\))³

A = \(\dfrac{8}{27}\) + \(\dfrac{1}{27}\)

A = \(\dfrac{9}{27}\)

A = \(\dfrac{1}{3}\) 

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

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

   \(=\left(-2y\right)^2-4y^2+4=4\)