Question

Por favor me ajudem nessas continhas estou desesperada

Answer 1

5) $1-\frac{4}{1} +\frac{4}{x^{2} } =0$

$1-4 +\frac{4}{x^{2} } =0$

$-3 +\frac{4}{x^{2} } =0$

$\frac{4}{x^{2} } =3$

$4=3x^{2} \\x^{2} =\frac{4}{3}$

$x =$ ± $\frac{2\sqrt{3} }{3}$

6) $x^{4} - 13x+36=0$

substituindo  $x^{2}$  por  $y$  

$y^{2} -13y+36=0\\y_{1} =4\\y_{2}= 9$

substituindo  $y$  por $x^{2}$

$x^{2} = 4\\x^{2} = 9\\\\x= \sqrt{4} \\x= \sqrt{9}$

$x$ ± $2$

$x$ ± $3$

$x_{1}= -3 \\x_{2}= -2\\x_{3}= 2\\x_{4}= 3$

7)

$\sqrt{3x+6}=x+2\\3x+6=(x+2)^{2} \\3(x+2)=(x+2)^{2} \\3(x+2)=(x+2)^{2}=0\\(x+2).(3-(x+2))=0\\(x+2).(3-x-2)=0\\(x+2).(1-x)=0\\x+2=0\\1-x=0\\x= -2\\x=1$

$x_{1} =-2\\x_{2} = 1$