Question

E(x) = 7x-3x^2/1-9x^2 - 3x/1-2x-3x^2 x ( 1 + 3x+x^2/ x+3)
a) descompuneti 1-2x-3x^2
b) arătați că E(x) = 4x/1+3x

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Answer 1

Răspuns:

Explicație pas cu pas:

a. -3x²-2x+1=

-(3x²+2x-1)=

-(3x²+3x-x-1)=

-[3x(x+1)-(x+1)]0=-(x+1)(3x-1)

b

. $E(x)=\frac{7x-3x^{2} }{1-9x^{2} } -\frac{3x}{1-2x-3x^{2} } \cdot (1+\frac{3x+x^{2} }{x+3} )\\E(x)=\frac{7x-3x^{2} }{(1-3x)(1+3x) } -\frac{3x}{-(x+1)(3x-1) } \cdot (1+\frac{x(3+x)}{x+3} )$

$\\E(x)=\frac{7x-3x^{2} }{(1-3x)(1+3x) } -\frac{3x}{-(x+1)(3x-1) } \cdot (1+x )$

$\\E(x)=\frac{7x-3x^{2} }{(1-3x)(1+3x) } +\frac{3x}{(3x-1) }$

$\\E(x)=\frac{7x-3x^{2}-3x-9x^{2} }{(1-3x)(1+3x) } =\frac{-12x^{2} +4x}{(1-3x)(1+3x)} =\frac{-4x(3x-1)}{(1-3x)(1+3x)}=\frac{4x}{1+3x}$

Answer 2

$\it a)\ 1-2x-3x^2=3-2-2x-3x^2=(3-3x^2)-2(1+x)=3(1^2-x^2)-2(1+x)=\\ \\ =3(1-x)(1+x)-2(1+x)=(1+x)(3-3x-2)=(1+x)(1-3x)$

$\it b)\ \ E(x)=\dfrac{7x-3x^2}{1-9x^2}-\dfrac{3x}{1-2x-3x^2}\cdot\Big(1+\dfrac{3x+x^2}{x+3}\Big)=\\ \\ \\ =\dfrac{7x-3x^2}{(1-3x)(1+3x)}-\dfrac{3x}{(1+x)(1-3x)}\cdot\Big(1+\dfrac{x(x+3)}{x+3}\Big)=\\ \\ \\ =\dfrac{7x-3x^2}{(1-3x)(1+3x)}-\dfrac{3x}{(1+x)(1-3x)}\cdot(1+x)=\dfrac{7x-3x^2}{(1-3x)(1+3x)}-\dfrac{^{1+3x)}3x}{1-3x}=\\ \\ \\ =\dfrac{7x-3x^2-3x-9x^2}{(1-3x)(1+3x)}=\dfrac{4x-12x^2}{(1-3x)(1+3x)}=\dfrac{4x(1-3x)}{(1-3x)(1+3x)}=\dfrac{4x}{1+3x}$