$\frac{a^{2}x-b^{2}x}{by-ay}=\frac{x(a^2-b^2)}{y(b-a)}=\frac{x(a+b)(a-b)}{y(b-a)}=\frac{-x(a+b)}{y}$

$\frac{x^2-5x-24}{x^2-64}=\frac{(x+3)(x-8)}{(x+8)(x-8)}=\frac{x+3}{x+8}$

$\frac{{\displaystyle \frac{a^3{b}^3}{x{y}^4}}}{{\displaystyle \frac{a^{2}b}{x^{2}y}}}=\frac{a^3{b}^3}{x{y}^4}·\frac{x^{2}y}{a^{2}b}=\frac{a{b}^{2}x}{y^3}$

$\frac{y^2+3y-40}{25-y^2}=\frac{(y-5)(y+8)}{(5-y)(5+y)}=-\frac{y+8}{y+5}$

$\frac{x^3+27}{3x+9}=\frac{(x+3)(x^2-3x+9)}{3(x+3)}=\frac{x^2-3x+9}{3}$

$\frac{c+d}{3c^2-3d^2}=\frac{c+d}{3(c^2-d^2)}=\frac{c+d}{3(c+d)(c-d)}=\frac{1}{3(c-d)}$

$\frac{{\displaystyle \frac{4x}{x+6}}}{{\displaystyle \frac{x^2-3x}{x^2+3x-18}}}=\frac{4x}{x+6}·\frac{x^2+3x-18}{x^2-3x}=\frac{4x}{x+6}·\frac{(x+6)(x-3)}{x(x-3)}=4$

$x^2-3x-28=0$

$(x+4)(x-7)=0$

إما $x+4=0$

$x=-4$

أو $x-7=0$

$x=7$

فالقيم التي تجعل العبارة غير معرفة هي: $-4 , 7$