Question

Buna! Ma poate ajuta cineva la urmatoarele 3 exercitii?

Answer 1

$4)4(2 - x) - 5(3x + 6)$

$= 8 - 4x - 15x - 30$

$= - 19x - 22$

$5)3 {x}^{3} - 27x = 3x( {x}^{2} - 9)$

$= 3x(x - 3)(x + 3)$

$6) \frac{ 3\sqrt{2} }{4 \sqrt{3} - 6} = \frac{3 \sqrt{2} (4 \sqrt{3} + 6)}{(4 \sqrt{3} - 6)(4 \sqrt{3} + 6) }$

$= \frac{12 \sqrt{6} + 18 \sqrt{2} }{ {(4 \sqrt{3}) }^{2} - {6}^{2} } = \frac{12 \sqrt{6} + 18 \sqrt{2} }{48 - 36}$

$= \frac{6(2 \sqrt{6} + 3 \sqrt{2} )}{12} = \frac{2 \sqrt{6} + 3 \sqrt{2} }{2}$

Answer 2

4)4(2-x)-5(3x+6)=

8-4x-15x-30=

(8-30)+(-4x-15x)=

-22-19x

$5) {3x}^{3} - 27x = \\ \\ 3x( \frac{ {3x}^{3} }{3x} + \frac{ - 27x}{3x}) = \\ \\ 3x( {x}^{2} - 9) = \\ \\ 3x( {x}^{2} - {3}^{2}) = \\ \\ 3x(x + 3)(x - 3)$

$6) \frac{3 \sqrt{2} }{4 \sqrt{3} - 6 } = \\ \\ \frac{12 \sqrt{6} + 18 \sqrt{2} }{ {(4 \sqrt{3}) }^{2} - {6}^{2} } = \\ \\ \frac{6(2 \sqrt{6} + 3 \sqrt{2}) }{ {4}^{2} { \sqrt{3} }^{2} - {6}^{2} } = \\ \\ \frac{6(2 \sqrt{6} + 3 \sqrt{2}) }{16 \times 3 - 36} = \\ \\ \frac{6(2 \sqrt{6} + 3 \sqrt{2}) }{48 - 36} = \\ \\ \frac{6(2 \sqrt{6} + 3 \sqrt{2}) }{12} = \\ \\ \frac{2 \sqrt{6} + 3 \sqrt{2} }{2}$