Resposta:

S = {-2, 4}

Explicação passo a passo:

[tex]\dfrac{5}{(x+1)^2} + \dfrac{4}{(x+1)} =1[/tex]

Inicialmente, tiramos o mmc que dará (x + 1)².

Pra facilitar, esse mmc equivale a simplesmente multiplicar todos os membros da equação por (x + 1)². Assim:

[tex]\dfrac{5}{(x+1)^2}\cdot (x+1)^2 + \dfrac{4}{(x+1)}\cdot (x+1)^2 =1\cdot (x+1)^2\to\\\\\\\\5 + 4(x+1)} =(x+1)^{2}\to 5 + 4x + 4 = x^{2}+2x+1\to\\\\\\\\4x+9=x^{2}+2x+1\to x^{2}+2x-4x+1-9\to\\\\\\ x^{2}-2x-8=0[/tex]

Agora, usamos a Fórmula de Bháskara para resolver a equação do segundo grau:

[tex]x=\dfrac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}\\\\\\\\x=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot 1\cdot (-8)}}{2\cdot 1}\\\\\\\\x=\dfrac{2\pm\sqrt{4+32}}{2}\\\\\\\\x=\dfrac{2\pm\sqrt{36}}{2}\\\\\\\\x=\dfrac{2\pm 6}{2}\\\\\\\\x'=\dfrac{2-6}{2}\to x'=-2\\\\\\x''=\dfrac{2+6}{2}\to x''=4\\\\\\S=\{-2,4\}[/tex]

Outra forma de fazer:

[tex]\dfrac{5}{(x+1)^2} + \dfrac{4}{(x+1)} =1[/tex]

Inicialmente, tiramos o mmc:

Lembrando que (x + 1)² = (x + 1)(x + 1). Assim, o mmc será (x + 1)(x + 1) ou (x + 1)²:

[tex]\dfrac{5}{(x+1)^2} + \dfrac{4}{(x+1)} =1\\\\\\\\\dfrac{5+4(x+1)=1(x+1)(x+1)}{(x+1)(x+1)}\\\\\\5+4(x+1)=(x+1)(x+1)\\\\\\5+4x+4=x^{2}+x+x+1\\\\\\4x+9=x^{2}+2x+1\\\\\\4x-2x-x^{2}+9-1=0\\\\\\2x-x^{2}+8=0\\\\\\-x^{2}+2x+8=0\:\:\:\:\: (-1)\\\\\\x^{2}-2x-8=0[/tex]

Agora é só seguir com a mesma resolução anterior usando a fórmula de Bháskara.