Answer
The solution is $a=1$
Work Step by Step
$\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a^{2}+a}$
Take out common factor $a$ from the denominator of the fraction on the right side of the equation:
$\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a(a+1)}$
Multiply the whole equation by $a(a+1)$:
$a(a+1)\Big[\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a(a+1)}\Big]$
$5(a+1)-7a=a^{2}-2a+4$
$5a+5-7a=a^{2}-2a+4$
Take all terms to the right side and simplify:
$0=a^{2}-2a+4-5a-5+7a$
$0=a^{2}-1$
Rearrange:
$a^{2}-1=0$
Take $1$ to the right side:
$a^{2}=1$
Take the square root of both sides:
$\sqrt{a^{2}}=\pm\sqrt{1}$
$a=\pm1$
The original equation is undefined for $a=-1$, so the solution is only $a=1$
