Question

eu imploro por favor eu tenho uma prova amanha!!!!!!!!!!!!!!!!!!!!!!!!!



1 RESOLVA AS SEGUINTES EXPRESOES FRACIONARIAS 

A)4/T+2 + 4/T-2= 2T/T²-4
B)1/Y+5 + 2/Y-5 = 7/Y²-25
C)5X-2/9X-X² + 3/X+3 - 1/3-X = 0

Answer 1

a)
$\frac{4}{t+2}+\frac{4}{t-2}=\frac{2t}{t^2-4}\\ \\ 4(t-2)+4(t+2)=2t\\ \\ 4t-8+4t+8=2t\\ \\ 6t=0\\ \\ \boxed{t=0}$

b)
$\frac{1}{y+5}+\frac{2}{y-5}=\frac{7}{y^2-25}\\ \\ y-5+2(y+5)=7\\ \\ y-5+2y+10=7\\ \\ 3y=7+5-10\\ \\ 3y=2\\ \\ \boxed{y=\frac{2}{3}}$

c)
$\frac{5x-2}{9x-x^2}+\frac{3}{x+3}-\frac{1}{3-x}=0\\ \\ 5x-2+3(3x-x^2)-3x-x^2=0\\ \\ 5x-2+9x-3x^2-3x-x^2=0\\ \\ -4x^2+11x-2=0\\ \\ \Delta=11^2-4.(-4)(-2)=121-32=89\\ \\ x=\frac{-11\pm\sqrt{89}}{-8}$

Answer 2

a) $\dfrac{4}{t+2}+\dfrac{4}{t-2}=\dfrac{2t}{t^2-4}$

$\dfrac{4}{t+2}+\dfrac{4}{t-2}=\dfrac{2t}{(t+2)(t-2)}$

$4(t-2)+4(t+2)=2t$

$4t-4+4t=4=2t$

$8t=2t$

$8t-2t=0~~\Rightarrow~~6t=0~~\Rightarrow~~t=0$.



b) $\dfrac{1}{y+5}+\dfrac{2}{y-5}=\dfrac{7}{y^2-25}$

$\dfrac{1}{y+5}+\dfrac{2}{y-5}=\dfrac{7}{(y+5)(y-5)}$

$y-5+2(y+5)=7$

$y-5+2y+10=7$

$3y+5=7~~\Rightarrow~~3y=2~~\Rightarrow~~y=\dfrac{2}{3}$.


c) $\dfrac{5x-2}{9-x^2}+\dfrac{3}{x+3}-\dfrac{1}{3-x}=0$

$\dfrac{5x-2}{(3+x)(3-x)}+\dfrac{3}{3+x}-\dfrac{1}{3-x}=0$

$5x-2+3(3-x)-1(x+3)=0$  

$5x-2+9-3x-x-3=0$

$x=3+2-9$

$x=-4$