Question

помогите только 6, 7 и 8​

Answer 1

$\frac{5x^2-10x^3+x^4}{5x^2} =\frac{x^2(5-10x+x^2)}{5x^2} = \frac{5-10x+x^2}{5} = \frac{x^2-10x+5}{5} \\ \\ \frac{5^2-10*5+5}{5} = \frac{25-50+5}{5} = \frac{-20}{5} = -4$

$\frac{12a^2b-3ab^2+ab}{3ab} = \frac{ab(12a-3b+1)}{3ab} = \frac{12a-3b+1}{3} \\\\\frac{12*5-3*0,2+1}{3} = \frac{60-0,6+1}{3} =\frac{60,4}{3} =\frac{302}{15}$

$\frac{4x^4+12x^3+x^2}{8x^2} =\frac{x^2(4x^2+12x+1)}{8x^2} = \frac{4x^2+12x+1}{8} \\ \frac{4*(-1)^2+12*(-1)+1}{8} =\frac{4-12+1}{8} =-\frac{7}{8}$