Question

Resolva: y/y-1 + 2y-1/y+1 = 6 (y+1) / 2y² - 2

Answer 1

Vamos lá:

$\frac{y}{y-1}+\:\frac{2y-1}{y+1}\:=\:\frac{6\:\left(y+1\right)}{2y^2-2}\\\frac{y\cdot \left(y+1\right)+\left(2y-1\right)\cdot \left(y-1\right)}{\left(y-1\right)\cdot \left(y+1\right)}=\:\frac{6y+6}{2\cdot \left(y^2-1\right)}\\\frac{\left(y^2+y\right)+\left(2y^2-2y-y+1\right)}{y^2-1^2}=\:\frac{6y+6}{2\cdot \left(y^2-1\right)}\\\frac{y^2+y+2y^2-2y-y+1}{y^2-1^2}=\:\frac{6y+6}{2\cdot \left(y^2-1\right)}\\3y^2-2y+1=\:\frac{6y+6}{2}\\2\cdot \left(3y^2-2y+1\right)=\:6y+6\\6y^2-4y+2=\:6y+6\\6y^2-4y-6y=6-2$
$6y^2-10y=4\\3y^2-5y=2\\3y^2-5y-2=0\\--------\\a=3\\b=-5\\c=-2\\--------\\\Delta \:=b^2-4ac\\\Delta \:=5^2-4\cdot 3\cdot \left(-2\right)\\\Delta \:=25+24\\\Delta \:=49\\--------\\y=\frac{-b\pm \sqrt{\Delta }}{2a}\\y=\frac{-\left(-5\right)\pm \sqrt{49}}{2\cdot 3}\\y=\frac{5\pm 7}{6}\\\\y'=\frac{5+7}{6}=2\\\\y''=\frac{5-7}{6}=-\frac{1}{3}$
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Espero ter ajudado!