Question

dau coroana ...................ex 2

Answer 1

$\displaystyle \it \frac{(x-3)^2}{3} - \frac{(1-x)(x+1)}{2} = \frac{(5x-1)(x+2)}{6} -3 \\ \\ \frac{x^2-2 \cdot x \cdot 3+3^2}{3} - \frac{1-x^2}{2} = \frac{5x^2+10x-x-2}{6} -3 \\ \\ \frac{x^2-6x+9}{3} - \frac{1-x^2}{2} = \frac{5x^2+9x-2}{6} -3 \\ \\ 2(x^2-6x+9)-3(1-x^2)=5x^2+9x-2-3 \cdot 6 \\ \\ 2x^2-12x+18-3+3x^2=5x^2+9x-2-18 \\ \\ 2x^2+3x^2-5x^2-12x-9x=-2-18-18+3 \\ \\ -21x=-35 \Rightarrow x= \frac{35}{21} \Rightarrow x= \frac{5}{3}$