Question

exercitiu de clasa a 7-a;
15 puncte!
coroana!

Rezolvati pe caiet, apoi trimiteti-mi poza!!!

Answer 1

   
[tex]\displaystyle\\ \bf\\ x\cdot \frac{x-2}{3}- \frac{1}{9}\cdot (x-4)^2-6=\left( \frac{x \sqrt{2} }{3}-1,(3) \right)\left( \frac{x \sqrt{2} }{3}+1,(3) \right)\\\\\\ \frac{x(x-2)}{3}- \frac{(x-4)^2}{9}-6=\left( \frac{x \sqrt{2} }{3}- \frac{13-1}{9} \right)\left( \frac{x \sqrt{2} }{3}+\frac{13-1}{9} \right)\\\\\\ \frac{x^2-2x}{3}- \frac{x^2-8x+16}{9}-6=\left( \frac{x \sqrt{2} }{3}- \frac{12}{9} \right)\left( \frac{x \sqrt{2} }{3}+\frac{12}{9} \right) [/tex]


[tex]\displaystyle\\ \bf\\ \frac{3(x^2-2x)}{3\cdot3}-\frac{x^2-8x+16}{9}-\frac{3\cdot6}{3}=\left(\frac{x \sqrt{2} }{3}-\frac{4}{3}\right)\left(\frac{x\sqrt{2}}{3}+\frac{4}{3}\right)\\\\\\ \frac{3x^2-6x}{9}-\frac{x^2-8x+16}{9}-\frac{18}{3}=\frac{x\sqrt{2}-4}{3}\cdot \frac{x\sqrt{2}+4}{3}\\\\\\ \frac{3x^2-6x-(x^2-8x+16)-18}{9}=\frac{\Big(x\sqrt{2}\Big)^2-\Big(4\Big)^2}{3^2}\\\\\\ \frac{3x^2-6x-x^2+8x-16-18}{9}=\frac{2x^2-16}{9}~~~\Big|\cdot9\\\\\\ 3x^2-6x-x^2+8x-16-18=2x^2-16 [/tex]

[tex]\displaystyle\\ \bf\\ \underbrace{3x^2-x^2 -2x^2}_{=0} \underbrace{-6x+8x}_{=2x} =\underbrace{-16+16+18}_{=18}\\\\ 2x = 18 ~~~\Longrightarrow ~~\text{Ecuatia data este de gradul 2 reductibila la gradul 1.}\\\\ \displaystyle\\ x = \frac{18}{2} = \boxed{9} [/tex]