Question

Resolva, em R, as equações:

a) x/(x-2) = 2x/(x+1)
b) 1/x + x/(x-2) = (3x+1)/x
c) 5 + √23x+31 = 3x
d) x + √x^2+1 = 3x + 1
e) 1/(x^2-4) = 2/(x+2) - 1/(x-2)

Answer 1

Explicação passo-a-passo:

a)

$\sf \dfrac{x}{x-2}=\dfrac{2x}{x+1}$

$\sf 2x\cdot(x-2)=x\cdot(x+1)$

$\sf 2x^2-4x=x^2+x$

$\sf 2x^2-x^2-4x-x=0$

$\sf x^2-5x=0$

$\sf x\cdot(x-5)=0$

• $\sf x'=0$

• $\sf x-5=0$

$\sf x"=5$

$\sf S=\{0,5\}$

b)

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

$\sf 1\cdot(x-2)+x\cdot x=(x-2)\cdot(3x+1)$

$\sf x-2+x^2=3x^2+x-6x-2$

$\sf x^2+x-2=3x^2-5x-2$

$\sf 3x^2-x^2-5x-x-2+2=0$

$\sf 2x^2-6x=0$

$\sf x^2-3x=0$

$\sf x\cdot(x-3)=0$

• $\sf x'=0$ (não serve)

• $\sf x-3=0$

$\sf x"=3$

$\sf S=\{3\}$

c)

$\sf 5+\sqrt{23x+31}=3x$

$\sf \sqrt{23x+31}=3x-5$

$\sf 23x+31=(3x-5)^2$

$\sf 23x+31=9x^2-30x+25$

$\sf 9x^2-30x+25-23x-31=0$

$\sf 9x^2-53x-6=0$

$\sf \Delta=(-53)^2-4\cdot9\cdot(-6)$

$\sf \Delta=2809+216$

$\sf \Delta=3025$

$\sf x=\dfrac{-(-53)\pm\sqrt{3025}}{2\cdot9}=\dfrac{53\pm55}{18}$

$\sf x'=\dfrac{53+55}{18}~\rightarrow~x'=\dfrac{108}{18}~\rightarrow~x'=6$

$\sf x"=\dfrac{53-55}{18}~\rightarrow~x"=\dfrac{-2}{18}~\rightarrow~x"=\dfrac{-1}{9}$ (não serve)

$\sf S=\{6\}$

d)

$\sf x+\sqrt{x^2+1}=3x+1$

$\sf \sqrt{x^2+1}=3x+1-x$

$\sf \sqrt{x^2+1}=2x+1$

$\sf x^2+1=(2x+1)^2$

$\sf x^2+1=4x^2+4x+1$

$\sf 4x^2+4x+1-x^2-1=0$

$\sf 3x^2+4x=0$

$\sf x\cdot(3x+4)=0$

$\sf x'=0$

• $\sf 3x+4=0$

$\sf 3x=-4$

$\sf x"=\dfrac{-4}{3}$ (não serve)

$\sf S=\{0\}$

e)

$\sf \dfrac{1}{x^2-4}=\dfrac{2}{x+2}-\dfrac{1}{x-2}$

$\sf 1=2\cdot(x-2)-(x+2)\cdot1$

$\sf 1=2x-4-x-2$

$\sf 1=x-6$

$\sf x=6+1$

$\sf x=7$

$\sf S=\{7\}$