Question

Va rog frumos exercitiul 4!

Answer 1

Am atasat rezolvarile

Answer 2

$a)3x + 3a + 6b = 3(x + a + 2b)$

$b)5x - 10y = 5(x - 2y)$

$c)4 {x}^{3} + 2 {x}^{2} - 6x = 2x(2 {x}^{2} + x - 3)$

$d)27 {x}^{3} {y}^{2} - 9 {x}^{2} {y}^{2} + 6 {x}^{2} y = 3 {x}^{2}y (9xy - 3y + 2)$

$e)2 \sqrt{3} {x}^{3} {y}^{6} - 5 \sqrt{3} {x}^{2} {y}^{4} + \sqrt{3} {x}^{3} {y}^{5} = \sqrt{3} {x}^{2} {y}^{4} (2x {y}^{2} - 5 + xy)$

$f) {a}^{2} x - {a}^{2} y = {a}^{2} (x - y)$

$g) \frac{1}{3} {x}^{2} - \frac{2}{9} x = \frac{1}{3} x( x - \frac{2}{3} )$

$h) {a}^{3} - ab - a = a( {a}^{2} - b - 1)$

$i) {x}^{3} {y}^{3} + {x}^{2} {y}^{2} + xy = xy( {x}^{2} {y}^{2} + xy + 1)$

$j) {a}^{2} x + a {x}^{2} + {a}^{2} {x}^{2} = ax(a + x + ax)$

$k)4 {a}^{3} {b}^{2} + 16 {a}^{2} {b}^{3} - 20 {a}^{3} {b}^{3} = 4 {a}^{2} {b}^{2} (a + 4b - 5ab)$