а)
[tex] \frac{1}{x - 1} = 2x \\ 2x(x - 1) = 1 \\ 2 {x}^{2} - 2x - 1 = 0 \\ d = ( - 2) {}^{2} - 4 \times 2 \times ( - 1) = \\ 4 + 8 = 12 \: ( d = 2 \sqrt{3} ) \\ x_{1} = \frac{2 + 2 \sqrt{3} }{2 \times 2} = \frac{2(1 + \sqrt{3} )}{2 \times 2} = \frac{1 + \sqrt{3} }{2} \\ x_{2} = \frac{2 - 2 \sqrt{3} }{2 \times 2} = \frac{2(1 - \sqrt{3} )}{2 \times 2} = \frac{1 - \sqrt{3} }{2} [/tex]
с)
[tex] \frac{v + 3}{v} + \frac{2}{3v} = 12 \\ \frac{3(v + 3) + 2}{3v} = 12 \\3v \times 12 = 3v + 9 + 2 \\ 36v - 3v = 11 \\ 33v = 11 \\ v = \frac{11}{33} \\ v = \frac{1}{3} [/tex]
е)
[tex] \frac{6y - 5}{2y} = \frac{7}{9} \\ 9(6y - 5) = 7 \times 2y \\ 54y - 45 = 14y \\ 54y - 14y = 45 \\ 40y = 45 \\ y = \frac{45}{40} \\ y = \frac{9}{8} \\ y = 1 \frac{1}{8} \\ y =1 .125[/tex]
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Answers & Comments
а)
[tex] \frac{1}{x - 1} = 2x \\ 2x(x - 1) = 1 \\ 2 {x}^{2} - 2x - 1 = 0 \\ d = ( - 2) {}^{2} - 4 \times 2 \times ( - 1) = \\ 4 + 8 = 12 \: ( d = 2 \sqrt{3} ) \\ x_{1} = \frac{2 + 2 \sqrt{3} }{2 \times 2} = \frac{2(1 + \sqrt{3} )}{2 \times 2} = \frac{1 + \sqrt{3} }{2} \\ x_{2} = \frac{2 - 2 \sqrt{3} }{2 \times 2} = \frac{2(1 - \sqrt{3} )}{2 \times 2} = \frac{1 - \sqrt{3} }{2} [/tex]
с)
[tex] \frac{v + 3}{v} + \frac{2}{3v} = 12 \\ \frac{3(v + 3) + 2}{3v} = 12 \\3v \times 12 = 3v + 9 + 2 \\ 36v - 3v = 11 \\ 33v = 11 \\ v = \frac{11}{33} \\ v = \frac{1}{3} [/tex]
е)
[tex] \frac{6y - 5}{2y} = \frac{7}{9} \\ 9(6y - 5) = 7 \times 2y \\ 54y - 45 = 14y \\ 54y - 14y = 45 \\ 40y = 45 \\ y = \frac{45}{40} \\ y = \frac{9}{8} \\ y = 1 \frac{1}{8} \\ y =1 .125[/tex]