Question

2x'2-5x+2/x'2 -10x+16
сократите дробь​

Answer 1

Смотри вложениеееееееееее

Answer 2

$\frac{2x^{2}-5x+2}{x^{2}-10x+16 }=\frac{2(x-2)(x-0,5)}{(x-2)(x-8)}=\frac{2(x-0,5)}{x-8}=\frac{2x-1}{x-8}$

$2x^{2}-5x+2=0\\\\D=(-5)^{2}-4*2*2=25-16=9=3^{2}\\\\x_{1}=\frac{5+3}{4}=2\\\\x_{2}=\frac{5-3}{4}=0,5\\\\2x^{2}-5x+2=2(x-2)(x-0,5)$