$4.$

$a)\dfrac{15-x^2}{-5x}=\dfrac{(...)}{5x}$ 

$⇒(15-x^2)5x=(...)(-5x)$

$⇒(-5x)[-(15-x^2)]=(...)(-5x)$

$⇒-(15-x^2)=(...)$

$⇒-15+x^2=(...)$

$b)\dfrac{x^2+8}{2x-1}=\dfrac{3x^3+24x}{(...)}$

$⇒(x^2+8)(...)=(3x^3+24x)(2x-1)$

$⇒(x^2+8)(...)=3x(x^2+8)(2x-1)$

$⇒(...)=3x(2x-1)$

$⇒(...)=6x^2-3x$

$5.$

$a)\dfrac{-x^2+2xy-y^2}{x+y}=\dfrac{(...)}{y^2-x^2}$ 

$⇒(-x^2+2xy-y^2)(y^2-x^2)=(x+y)(...)$

$⇒(-x^2+2xy-y^2)(y-x)(y+x)=(x+y)(...)$

$⇒(-x^2+2xy-y^2)(y-x)=(...)$

$⇒-x^2y+x^3+2xy^2-2x^2y-y^3+xy^2=(...)$

$⇒x^3-3x^2y+3xy^2-y^3=(...)$

$⇒(x-y)^3=(...)$

$a)\dfrac{x^5-1}{x^2-1}=\dfrac{(...)}{x+1}$ 

$⇒(x^5-1)(x+1)=(x^2-1)(...)$

$⇒(x^5-1)(x+1)=(x-1)(x+1)(...)$

$⇒(x^5-1)=(x-1)(...)$

$⇒(x^5-1):(x-1)=(...)$

$⇒(...)=x^4+x^3+x^2+x+1$

#andy

\[\begin{array}{l}
15)\\
a)\dfrac{{15 - {x^2}}}{{ - 5x}} = \dfrac{?}{{5x}}\\
 \Leftrightarrow (15 - {x^2})5x = ?.( - 5x)\\
 \Leftrightarrow ( - 5x)[ - (15 - {x^2})] = ?.( - 5x)\\
 \Leftrightarrow  - (15 - {x^2}) = ?\\
 \Leftrightarrow  - 15 + {x^2} = ?\\
b)\dfrac{{{x^2} + 8}}{{2x - 1}} = \dfrac{{3{x^3} + 24x}}{?}\\
 \Leftrightarrow ({x^2} + 8).? = (3{x^3} + 24x)(2x - 1)\\
 \Leftrightarrow ({x^2} + 8).? = 3x({x^2} + 8)(2x - 1)\\
 \Leftrightarrow ? = 3x(2x - 1)\\
 \Leftrightarrow ? = 6{x^2} - 3x\\
16)\\
a)\dfrac{{ - {x^2} + 2xy - {y^2}}}{{x + y}} = \dfrac{?}{{{y^2} - {x^2}}}\\
 \Leftrightarrow ( - {x^2} + 2xy - {y^2})({y^2} - {x^2}) = (x + y).?\\
 \Leftrightarrow ( - {x^2} + 2xy - {y^2})(y - x)(y + x) = (x + y).?\\
 \Leftrightarrow ( - {x^2} + 2xy - {y^2})(y - x) = ?\\
 \Leftrightarrow  - {x^2}y + {x^3} + 2x{y^2} - 2{x^2}y - {y^3} + x{y^2} = ?\\
 \Leftrightarrow {x^3} - 3{x^2}y + 3x{y^2} - {y^3} = ?\\
 \Leftrightarrow {(x - y)^3} = ?\\
b)\dfrac{{{x^5} - 1}}{{{x^2} - 1}} = \dfrac{?}{{x + 1}}\\
 \Leftrightarrow ({x^5} - 1)(x + 1) = ({x^2} - 1).?\\
 \Leftrightarrow ({x^5} - 1)(x + 1) = (x - 1)(x + 1).?\\
 \Leftrightarrow ({x^5} - 1) = (x - 1).?\\
 \Leftrightarrow ({x^5} - 1):(x - 1) = ?\\
 \Leftrightarrow ? = {x^4} + {x^3} + {x^2} + x + 1
\end{array}\]