Question

Resolva a expressão algébrica,
Simplificando o resultado se necessario.

   b   \\\\+\\a\\\-\\\3a 
ab+b\\\\\\\\1\\\\\\a+1

Answer 1

$\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{b}{b(a+1)}+\dfrac{a}{1}-\dfrac{3a}{a+1}\\\\\\\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{1}{a+1}+\dfrac{a(a+1)}{(a+1)}-\dfrac{3a}{a+1}\\\\\\\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{1}{a+1}+\dfrac{a^{2}+a}{a+1}-\dfrac{3a}{a+1}$

Denominadores iguais, somamos os numeradores e conservamos o denominador:

$\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{1+a^{2}+a-3a}{a+1}\\\\\\\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{a^{2}-2a+1}{a+1}\\\\\\\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{a^{2}-2\cdot a\cdot1+1^{2}}{a+1}$

Veja que a estrutura do numerador é a estrutura do resultado de um produto notável: O quadrado da diferença de 2 termos

$a^{2}-2a+1=(a-1)^{2}$

Logo:

$\boxed{\boxed{\dfrac{b}{ab+b}+\dfrac{a}{1}-\dfrac{3a}{a+1}=\dfrac{(a-1)^{2}}{a+1}}}$

Answer 2

$\frac{b}{ab+b}+\frac{a}{1}-\frac{3a}{a+1}=\\\\\frac{b}{b(a+1)}+a-\frac{3a}{a+1}=\\\\\frac{1}{a+1}-\frac{3a}{a+1}+a=\\\\\frac{1-3a}{a+1}+a=\\\\\frac{1-3a+a(a+1)}{a+1}=\\\\\frac{1-3a+a^2+a}{a+1}=\\\\\frac{a^2-2a+1}{a+1}=\\\\\boxed{\frac{(a-1)^2}{a+1}}$