Question

$E4(x)=\frac{2x^{2}+16x+32}{3x^{2}-48}:[(\frac{2}{x+2}+\frac{2}{6-3x}+\frac{8}{x^{2}-4}):\frac{2x-8}{x^{2}-3x+2}]$

x€|R \ {-4; -2; 1; 2; 4}

a)Demonstraţi că $E4(x)=\frac{x+4}{x-1}$

Answer 1

Rezolvarea este în atașament.

Answer 2

$\frac{2( {x}^{2} + 8x + 16) }{3( {x}^{2} - 16) } \div (( \frac{2}{x + 2} + \frac{2}{3(2 - x)} + \frac{8}{(x - 2)(x + 2)} ) \times \frac{ {x}^{2} - 3x + 2 }{2x - 8}$
$\frac{2 {(x + 4)}^{2} }{3(x - 4)(x + 4)} \div (( \frac{2}{x + 2} + \frac{2}{3( - (x - 2))} + \frac{8}{(x - 2)(x + 2)} ) \times \frac{ {x}^{2} - x - 2x + 2 }{2(x - 4)} )$
$\frac{2(x + 4)}{3(x - 4)} \div (( \frac{2}{x + 2} - \frac{2}{3(x - 2)} + \frac{8}{(x - 2)(x + 2)} ) \times \frac{x(x - 1) - 2(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div ( \frac{6(x - 2) - 2(x + 2) + 24}{3(x - 2)(x + 2)} \times \frac{(x - 2)(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div ( \frac{6x - 12 - 2x - 4 + 24}{3(x - 2)(x + 2)} \times \frac{(x - 2)(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div ( \frac{4x + 8}{3(x - 2)(x + 2)} \times \frac{(x - 2)(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div ( \frac{4(x + 2)}{3(x -2)(x + 2)} \times \frac{(x - 2)(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div \frac{4}{3(x - 2)} \times \frac{(x - 2)(x - 1)}{2(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div ( \frac{2}{3} \times \frac{x - 1}{x - 4} )$
$\frac{2(x + 4)}{3(x - 4)} \div \frac{2(x - 1)}{3(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \div \frac{2x - 2}{3(x - 4)}$
$\frac{2(x + 4)}{3(x - 4)} \times \frac{3(x - 4)}{2x - 2}$
$2(x + 4) \times \frac{1}{2(x - 1)}$
$(x + 4) \times \frac{1}{x - 1}$
$\frac{x + 4}{x - 1}$