Respuesta :
First take
[tex]\frac{x^2+3x-70}{x^2+2x-80}\ldots\ldots(1)[/tex][tex]\begin{gathered} x^2+3x-70=x^2-7x+10x-70 \\ =x\mleft(x-7\mright)+10\mleft(x-7\mright) \\ =\mleft(x-7\mright)\mleft(x+10\mright)\ldots\text{.}\mathrm{}(2) \end{gathered}[/tex][tex]\begin{gathered} x^2+2x-80=x^2-8x+10x-80 \\ =x(x-8)+10(x-8) \\ =(x-8)(x+10)\ldots\ldots\text{.}(3) \end{gathered}[/tex]Put the values of (2) and (3) in equation (1), it becomes,
[tex]\begin{gathered} \frac{x^2+3x-70}{x^2+2x-80}=\frac{(x-7)(x+10)}{(x-8)(x+10)} \\ =\frac{x-7}{x-8}\ldots\ldots(4) \end{gathered}[/tex][tex]\frac{x^2+2x-80}{x^2+6x-16}\ldots\text{.}(5)[/tex]From equation (3)
[tex]x^2+2x-80=(x-8)(x-10)[/tex][tex]\begin{gathered} x^2+6x-16=x^2-2x+8x-16 \\ =x(x-2)+8(x-2) \\ =(x-2)(x+8)\ldots\ldots\text{.}\mathrm{}(6) \end{gathered}[/tex][tex]\frac{x^2+2x-80}{x^2+6x-16}=\frac{(x-8)(x+10)}{(x-2)(x+8)}\ldots\text{.}(7)[/tex]From equation (4) and (7),
[tex]\begin{gathered} \frac{\frac{x^2+3x-70}{x^2+2x-80}}{\frac{x^2+2x-80}{x^2+6x-16}}=\frac{\frac{x-7}{x-8}}{\frac{(x-8)(x+10)}{(x-2)(x+8)}} \\ =\frac{x-7}{x-8}\cdot\frac{(x-2)(x+8)}{(x-8)(x+10)} \\ =\frac{(x-7)(x-2)(x+8)}{(x-8)^2(x+10)} \end{gathered}[/tex]Otras preguntas
