Question

e)x²-1/3=1/6x²
f)x²/4+1/10=x²/5+x/2
g)x+6=4x/x-2(x≠-2)
h)2x/x-3=x+1/x+3(x≠-3,x≠3)
i)x+1/x-1 x+1/(x≠1,×≠1) =x-3x²/x²-1..

me ajudem ........
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Answer 1

${x}^{2} - \frac{1}{3} = \frac{1}{6} {x}^{2} \\ \\ 6 {x}^{2} - 2 = x {}^{2} \\ 6 {x}^{2} - {x}^{2} = 2 \\ 5 {x}^{2} = 2 \\ {x}^{2} = \frac{2}{5} \\ x = + - \frac{ \sqrt{10} }{5} \ \\ x1 = + \frac{ \sqrt{10} }{5} \\ x2 = - \frac{ \sqrt{10} }{5}$

$\frac{ {x}^{2} }{4} + \frac{1}{10} = \frac{ {x}^{2} }{5} + \frac{x}{2} \\ x1 = 5 - \sqrt{23} \\ x2 = 5 + \sqrt{23}$

$x + 6 = \frac{4x}{x - 2} \\ x1 = - 2 \sqrt{3} \\ x2 = 2 \sqrt{3}$

$\frac{2x}{x - 3} = \frac{x + 1}{x + 3} \\ \\ \\ x1 = - 4 - \sqrt{13} \\ x2 = - 4 + \sqrt{13}$

$\frac{x}{x - 1} + \frac{1}{x + 1} = \frac{x - 3 {x}^{2} }{ {x}^{2} - 1 } \\ x1 = \frac{ - 1 \sqrt{17} }{8} \\ x2 = \frac{ - 1 + \sqrt{17 } }{8}$