Question

x/x-2 + 4/x-1 =5 resolução

Answer 1

$\large\begin{array}{l} \mathsf{\dfrac{x}{x-2}+\dfrac{4}{x-1}=5}\\\\\\ \textsf{Reduza as fra\c{c}\~oes do lado esquerdo ao mesmo denominador:}\\\\ \mathsf{\dfrac{x(x-1)}{(x-2)(x-1)}+\dfrac{4(x-2)}{(x-2)(x-1)}=5}\\\\ \mathsf{\dfrac{x(x-1)+4(x-2)}{(x-2)(x-1)}=5} \end{array}$


$\large\begin{array}{l} \textsf{Agora, aplique a distributiva para eliminar os par\^enteses:}\\\\ \mathsf{\dfrac{x^2-x+4x-8}{x^2-x-2x+2}=5}\\\\ \mathsf{\dfrac{x^2+3x-8}{x^2-3x+2}=5}\\\\ \mathsf{x^2+3x-8=5(x^2-3x+2)}\\\\ \mathsf{x^2+3x-8=5x^2-15x+10}\\\\ \mathsf{0=5x^2-15x+10-x^2-3x+8} \end{array}$

$\large\begin{array}{l} \mathsf{4x^2-18x+18=0}\\\\ \mathsf{2\cdot (2x^2-9x+9)=0}\\\\ \mathsf{2x^2-9x+9=0}\quad\Rightarrow\quad\left\{ \!\begin{array}{l}\mathsf{a=2}\\\mathsf{b=-9}\\\mathsf{c=9} \end{array} \right. \end{array}$


$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-9)^2-4\cdot 2\cdot 9}\\\\ \mathsf{\Delta=81-72}\\\\ \mathsf{\Delta=9}\\\\ \mathsf{\Delta=3^2} \end{array}$


$\large\begin{array}{l} \mathsf{x=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{x=\dfrac{-(-9)\pm \sqrt{3^2}}{2\cdot 2}}\\\\ \mathsf{x=\dfrac{9\pm 3}{4}}\\\\ \begin{array}{rcl} \mathsf{x=\dfrac{9-3}{4}}&~\textsf{ ou }~&\mathsf{x=\dfrac{9+3}{4}}\\\\ \mathsf{x=\dfrac{6}{4}}&~\textsf{ ou }~&\mathsf{x=\dfrac{12}{4}}\\\\ \end{array}\\ \quad~\boxed{\begin{array}{rcl} \mathsf{x=\dfrac{3}{2}}&~\textsf{ ou }~&\mathsf{x=3} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{Deve-se ter aten\c{c}\~ao ao fato de estes valores n\~ao anularem os}\\ \textsf{denominadores das fra\c{c}\~oes envolvidas na equa\c{c}\~ao dada inicialmente}\\ \textsf{(verificar as condi\c{c}\~oes de exist\^encia).}\\ \\\textsf{Os dois valores encontrados para x s\~ao solu\c{c}\~oes para a equa\c{c}\~ao.}\\\\ \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\left\{\dfrac{3}{2},\,3\right\}.}\\ \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$