Question

Дам 30 балів Все на фото​

Answer 1

Ответ:

$1)\ \ x^2-3x-18=(x-6)(x+3)\ \ ,\ \ \ x^2+5x-14=(x-2)(x+7)\\\\-x^2+3x+4=-(x^2-3x-4)=-(x-4)(x+1)\\\\5x^2+8x-4=5(x+2)(x-\frac{2}{5})=(x+2)(5x-2)\\\\2a^2-3a-1=2(a-\frac{3-\sqrt{17}}{4})(a-\frac{3+\sqrt{17}}{4})\\\\4b^2-11b-3=4(b-3)(b+\frac{1}{4})=(b-3)(4b+1)\\\\-\dfrac{1}{4}\, x^2-2x-3=-\dfrac{1}{4}\, (x^2+8x+12)=-\dfrac{1}{4}\, (x+6)(x+2)$

$0,3m^2-3x+7,5=0,3\, (m^2-10m+25)=0,3(m-5)^2\\\\-0,5x^2+x+1,5=-0,5(x^2-2x-3)=-0,5(x-3)(x+1)$

$2)\ \ \dfrac{2x+12}{x^2+3x-18}=\dfrac{2(x+6)}{(x+6)(x-3)}=\dfrac{2}{x-3}\\\\\dfrac{x^2+9x+14}{x^2+7x}=\dfrac{(x+7)(x+2)}{x(x+7)}=\dfrac{x+2}{x}\\\\\dfrac{4x^2+x-3}{x^2-1}=\dfrac{(4x-3)(x+1)}{(x-1)(x+1)}=\dfrac{4x-3}{x-1}$

$\dfrac{2y^2+3y-5}{y^2-2y+1}=\dfrac{(y-1)(2y+5)}{(y-1)^2}=\dfrac{2y+5}{y-1}\\\\\dfrac{a^2+5a+4}{a^2-a-20}=\dfrac{(a+1)(a+4)}{(a-5)(a+4)}=\dfrac{a+1}{a-5}\\\\\dfrac{3+20b-7b^2}{7b^2-6b-1}=\dfrac{-(7b+1)(b-3)}{(7b+1)(b-1)}=\dfrac{3-b}{b-1}$