Question

ajutatima urgent rezolvare pe lung

Answer 1

Răspuns:

Explicație pas cu pas:

$a)~E(x)=(1+\frac{5x}{1-3x} ):\frac{1-4x^{2}}{1-6x+9x^{2}}=\frac{1-3x+5x}{1-3x}:\frac{1^{2}-(2x)^{2}}{1^{2}-2*1*3x+(3x)^{2}}=\frac{1+2x}{1-3x} :\frac{(1-2x)(1+2x)}{(1-3x)^{2}}= \frac{1+2x}{1-3x} *\frac{(1-3x)^{2}}{(1-2x)(1+2x)}=\frac{(1+2x)(1-3x)^{2}}{(1-3x)(1-2x)(1+2x)} =\frac{1-3x}{1-2x}.\\Deci~E(x)=\frac{1-3x}{1-2x}.$

b) DVA se afla din conditia 1-2x≠0, deci -2x≠-1, ⇒x≠(-1):(-2), ⇒x≠1/2.

DVA=R\{1/2}.