5x-4=x+2 

=>5x-x=2+4

=>4x=6

=> x=6/4=1,5

Theo bài ra , ta có : 

\(5x-4=x+2\)

\(\Rightarrow5x-x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)

- xy - xy - xy = - 1xy - 1xy - 1xy = - 3xy

X³ -XY - X³ - X²Y + X²Y -XY = -2XY

\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\times\frac{x^2+8x+16}{32}\)

ĐKXĐ : \(x\ne\pm4\)

\(=\left(\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\times\frac{\left(x+4\right)^2}{32}\)

\(=\left(\frac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\right)\times\frac{\left(x+4\right)^2}{32}\)

\(=\frac{32}{\left(x-4\right)\left(x+4\right)}\times\frac{\left(x+4\right)^2}{32}\)

\(=\frac{x+4}{x-4}\)

\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right).\frac{x^2+8x+16}{32}\)

\(=\left(\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right).\frac{\left(x+4\right)^2}{32}\)

\(=\frac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}.\frac{\left(x+4\right)^2}{32}=\frac{32}{\left(x-4\right)\left(x+4\right)}.\frac{\left(x+4\right)^2}{32}=\frac{x+4}{x-4}\)

(x-2)(x+2)(x+4) = (x^2 - 4) (x+4) = x^3 + 4x^2 -4x - 16 

\(=\dfrac{x^2-4y}{xy}\cdot\dfrac{x^2}{x-y}=\dfrac{x\left(x^2-4y\right)}{y\left(x-y\right)}\)

x^n-2( x³ + x² - 5 )

= xⁿ⁺¹ + xⁿ - 5x^n-2