• Matéria: Matemática
  • Autor: Hungrianne0985
  • Perguntado 8 anos atrás

Resolva as equações fracionarias.

a) 2/x-3 - 1/4 = 5/x-3 - 1/3

b) x+4/x+9 = x/ x+3

Me ajudeemmmmm


Dunskyl: É 2 sobre x ou 2 sobre x-3 ?
Hungrianne0985: 2 sobre x-3

Respostas

respondido por: Dunskyl
7
a)

 \frac{2}{x-3}  -  \frac{1}{4}  =  \frac{5}{x-3}  -  \frac{1}{3}  \\  \\  \frac{2\cdot4-1\cdot(x-3)}{(x-3)\cdot4} = \frac{5\cdot3-1\cdot(x-3)}{(x-3)\cdot3}

\frac{8-x+3}{4x-12} = \frac{15-x+3}{3x-9} \\  \\  (8-x+3)\cdot(3x-9)=(15-x+3)\cdot(4x-12) \\  \\  (11-x)\cdot(3x-9)=(18-x)\cdot(4x-12) \\  \\ 33x-99-3x^{2}+9x=72x-216-4x^{2}+12x

4x^{2}-3x^{2}+33x-12x-72x+9x-99+216=0 \\  \\ x^{2}-42x+117=0

\Delta=(-42)^{2}-4(1)(117)=1764-468=1296

x= \frac{-(-42)\pm \sqrt{1296} }{2}  \\  \\ x= \frac{42\pm36}{2}

x'= \frac{42+36}{2}  \\  x'= \frac{78}{2}  \\  x'=39

x''= \frac{42-36}{2}  \\ x''= \frac{6}{2}  \\ x''= 3

b)

 \frac{x+4}{x+9}  =  \frac{x}{x+3}  \\  \\ (x+4)\cdot(x+3)=x\cdot(x+9) \\  \\ x^{2}+3x+4x+12=x^{2}+9x \\  \\ x^{2}-x^{2}+3x+4x-9x=-12 \\  \\ -2x=-12 \\  \\ x=6
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