Question

Trabalho sobre equações!

Answer 1

Primeira figura:

a) $\dfrac{4}{1-x}-\dfrac{5}{x+1}=\dfrac{3x-2}{1-x^2}$.

$\dfrac{4(x+1)}{(1-x)(x+1)}-\dfrac{5(1-x)}{(1-x)(x+1)}=\dfrac{3x-2}{(1-x)(x+1)}$

$4(x+1)-5(1-x)=3x-2$

$4x+4-5+5x=3x-2$

$4x+5x-3x=5-4-2$

$6x=-1$

$x=\dfrac{-1}{6}$


b) $\dfrac{1}{x+4}+\dfrac{1}{x-4}=\dfrac{-8}{x^2-16}$

$\dfrac{(x-4)+(x+4)}{(x+4)(x-4)}=\dfrac{-8}{(x+4)(x-4)}$

$(x-4)+(x+4)=-8$

$2x=-8$

$x=-4$

Mas, note que, se $x=-4$ o denominador da primeira fração será zero.

Portanto, esta equação não possui solução.

c) $\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{2}{x+2}=\dfrac{3}{x^2-4}$

$\dfrac{(x-2)(x+2)+x(x+2)-2x(x-2)}{x(x-2)(x+2)}=\dfrac{3}{(x-2)(x+2)}$

$(x-2)(x+2)+x(x+2)-2x(x-2)=3x$

$x^2-4+x^2+2x-2x^2+4x=3x$

$6x-4=3x$

$3x=4$

$x=\dfrac{4}{3}$

d) $\dfrac{3}{x}+\dfrac{1}{x-1}=\dfrac{-3x+4}{x^2-x}$

$\dfrac{3(x-1)+x}{x(x-1)}=\dfrac{-3x+4}{x(x-1)}$

$3(x-1)+x=-3x+4$

$3x-3+x=-3x+4$

$3x+x+3x=4+3$

$7x=7$

$x=1$.


Segunda figura:

a) $\dfrac{x-1}{a+1}+\dfrac{x+1}{a+3}=2$

$\dfrac{(x-1)(a+3)+(x+1)(a+1)}{(a+1)(a+3)}=2$

$ax+3x-a-3+ax+x+a+1=2(a+1)(a+3)$

$x(a+3+a+1)=2(a+1)(a+3)+8$

$x(2a+4)=(2a+4)(a+2)$

$x=\dfrac{(2a+4)(a+2)}{(2a+4)}=a+2$


b) $a^2x+10a+25x+a^2+25$ 

$a^2x-25x=a^2-10a+25$

$x(a^2-25)=(a-5)^2$

$x(a-5)(a+5)=(a-5)(a-5)$

$x=\dfrac{(a-5)(a-5)}{(a-5)(a+5)}=\dfrac{a-5}{a+5}$.


c) $\dfrac{a}{x+a}+\dfrac{ax}{x^2-a^2}=\dfrac{a}{x-a}$.

$\dfrac{a(x-a)+ax}{(x+a)(x-a)}=\dfrac{a}{x-a}$

$a(x-a)+ax=a(x+a)$

$ax-a^2+ax=ax+a^2$

$ax=2a^2$

$x=\dfrac{2a^2}{a}=2a$


Terceira figura:

$\dfrac{210}{x}=\dfrac{120}{x-3}$

$210(x-3)=120x$

$210x-120x=210\cdot3$

$90x=630$

$x=\dfrac{630}{90}=7$

Quarta figura:

$\dfrac{360}{x}=\dfrac{270}{x-2}$

$360(x-2)=270x$

$360x-270x=360\cdot2$

$90x=720$

$x=\dfrac{720}{90}=8$