Question

me podrian ayudar x fas lo de la foto es urgente

Answer 1

$\frac{\left(3x-1\right)^2}{x-1}=\frac{18x-1}{2}$
$\left(3x-1\right)^2\cdot \:2=\left(x-1\right)\left(18x-1\right)$

$\left(3x-1\right)^2\cdot \:2=2\left(9x^2-6x+1\right)=2\cdot \:9x^2+2\left(-6x\right)+2\cdot \:1$
$=2\cdot \:9x^2-2\cdot \:6x+2\cdot \:1=18x^2-12x+2$


$\left(x-1\right)\left(18x-1\right)=x\cdot \:18x+x\left(-1\right)-1\cdot \:18x-1\cdot \left(-1\right)$
$=x\cdot \:18x-x\cdot \:1-1\cdot \:18x+1\cdot \:1=18x^2-19x+1$


$18x^2-12x+2=18x^2-19x+1$
$18x^2-12x+2-2=18x^2-19x+1-2$
$18x^2-12x=18x^2-19x-1$
$18x^2-12x-\left(18x^2-19x\right)=18x^2-19x-1-\left(18x^2-19x\right)$
$7x=-1$
$\frac{7x}{7}=\frac{-1}{7}$
$x=-\frac{1}{7}$