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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

Giải:

a) \(3x^2y-6xy^2\)

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

Vậy ...

b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)

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

\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

Vậy ...

c) \(25a^2-c^2\)

\(=\left(5a-c\right)\left(5a+c\right)\)

Vậy ...

d) \(4-36x+81x^2\)

\(=2^2-2.2.9x+\left(9x\right)^2\)

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

Vậy ...

e) \(\left(x+7\right)2-\left(2x-9\right)2\)

\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)

\(=2\left(x+7-2x+9\right)\)

\(=2\left(16-x\right)\)

Vậy ...

f) \(x^2-6x+8\)

\(=x^2-6x+9-1\)

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

\(=\left(x-4\right)\left(x-2\right)\)

Vậy ...

a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)

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

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

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

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

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

\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)

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

\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)

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

Viết đề rõ chút chứ nhìn ko ra

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)

=(x+1)(x^4+x^3+x^2+x+1)

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