Question

Сократите дробь: 8x2-2x-1/16x2+8x+1

Answer 1

Ответ:

$\frac{2x-1}{4x+1}$

Объяснение:

$\frac{8x^{2}-2x-1}{16x^{2}+8x+1};$

$8x^{2}-2x-1=0;$

$D=b^{2}-4*a*c;$

$D=(-2)^{2}-4*8*(-1)=4+32=36;$

$x_{1}=\frac{-b+\sqrt{D}}{2a};$

$x_{1}=\frac{-(-2)+\sqrt{36}}{2*8}=\frac{2+6}{16}=\frac{8}{16}=\frac{1}{2}=0,5;$

$x_{2}=\frac{-b-\sqrt{D}}{2a};$

$x_{2}=\frac{-(-2)-\sqrt{36}}{2*8}=\frac{2-6}{16}=\frac{-4}{16}=-\frac{1}{4}=-0,25;$

$ax^{2}+bx+c=a(x-x_{1})(x-x_{2});$

$8x^{2}-2x-1=8*(x-0,5)*(x+0,25)=4*(2x-1)(x+0,25)=(2x-1)(4x+1);$

$16x^{2}+8x+1=0;$

$D=b^{2}-4*a*c;$

$D=8^{2}-4*16*1=64-64=0;$

$x=\frac{-b}{2a};$

$x=\frac{-8}{2*16}=\frac{-8}{32}=-\frac{1}{4}=-0,25;$

$16x^{2}+8x+1=16*(x+0,25)*(x+0,25)=(4x+1)(4x+1);$

$\frac{8x^{2}-2x-1}{16x^{2}+8x+1}=\frac{(2x-1)(4x+1)}{(4x+1)(4x+1)}=\frac{2x-1}{4x+1};$