Question

Применяя метод выделения полного квадрата, разложите числитель и знаменатель дроби на множители и сократите дробь.

Answer 1

$\dfrac{ 8x^{2} - 2x - 1 }{-6 x^{2} + 5x - 1}$

$8x^{2} - 2x - 1 =8(x^2- \frac{1}{4} x- \frac{1}{8} )=8(x^2-2\cdot x\cdot \frac{1}{8} +( \frac{1}{8} )^2-( \frac{1}{8} )^2- \frac{1}{8} )= \\\ =8((x-\frac{1}{8} )^2- \frac{1}{64} - \frac{8}{64} )= 8((x-\frac{1}{8} )^2 - \frac{9}{64} )=8((x-\frac{1}{8} )^2 - (\frac{3}{8})^2 )= \\\ =8(x-\frac{1}{8} -\frac{3}{8})(x-\frac{1}{8} +\frac{3}{8})= 8(x-\frac{4}{8} )(x+\frac{2}{8})=8(x-\frac{1}{2} )(x+\frac{1}{4})$

$-6 x^{2} + 5x - 1=-6(x^2- \frac{5}{6} x+ \frac{1}{6} )= \\\ =-6(x^2- 2\cdot x\cdot \frac{5}{12}+( \frac{5}{12} )^2-( \frac{5}{12} )^2+ \frac{1}{6} )= \\\ =-6((x- \frac{5}{12} )^2- \frac{25}{144} + \frac{24}{144} )= -6((x- \frac{5}{12} )^2- \frac{1}{144} )= \\\ =-6((x- \frac{5}{12} )^2- (\frac{1}{12})^2 )= -6(x- \frac{5}{12} -\frac{1}{12})(x- \frac{5}{12} +\frac{1}{12})= \\\ =-6(x- \frac{6}{12})(x- \frac{4}{12} )= -6(x- \frac{1}{2})(x- \frac{1}{3} )$

$\dfrac{ 8x^{2} - 2x - 1 }{-6 x^{2} + 5x - 1} = \dfrac{ 8(x-\frac{1}{2} )(x+\frac{1}{4})}{-6(x- \frac{1}{2})(x- \frac{1}{3} )} = -\dfrac{ 4(x+\frac{1}{4})}{3(x- \frac{1}{3} )} =-\dfrac{ 4x+1}{3x- 1}$