1. Найдите произведение корней уравнения: х3+6x2+4x-5 ____________ = 1 x2+8x+15 2....

Question 1

1. Найдите произведение корней уравнения:

х3+6x2+4x-5

____________ = 1

x2+8x+15

2. Вычислите дроби:

4x2+xy-2xz y x

__________ при условии, что ___ = 2, ___ = 3

3y2+2xz-z2 z y

Question 2

1.
$\dfrac{x^3+6x^2+4x-5}{x^2+8x+15}=1\\ \dfrac{(x^3+6x^2+4x-5)-(x^2+8x+15)}{(x+3)(x+5)}=0 \\ O.D.3.: x \neq-3,\ x \neq-5\\ x^3+5x^2-4x-20=0\\ x^2(x+5)-4(x+5)=0\\ (x+5)(x-2)(x+2)=0\\ x_1=-5,\ x_2=-2,\ x_3=2$
C учетом О.Д.З. получим х=-2, х=2.
Ответ: -2; 2.
2.
y=2z\\\\ \dfrac{x}{y}=3\ => x=3y=3(2z)=6z\\\\ \dfrac{4x^2+xy-2xz}{3y^2+2xz-z^2}=\dfrac{4(6z)^2+6z*2z-2z*6z}{3(2z)^2+2z*6z-z^2}=\\\\ =\dfrac{144z^2+12z^2-12z^2}{12z^2+12z^2-z^2}=\dfrac{144z^2}{23z^2}=\dfrac{144}{23}=6\frac{6}{23}" alt="\dfrac{y}{z}=2\ => y=2z\\\\ \dfrac{x}{y}=3\ => x=3y=3(2z)=6z\\\\ \dfrac{4x^2+xy-2xz}{3y^2+2xz-z^2}=\dfrac{4(6z)^2+6z*2z-2z*6z}{3(2z)^2+2z*6z-z^2}=\\\\ =\dfrac{144z^2+12z^2-12z^2}{12z^2+12z^2-z^2}=\dfrac{144z^2}{23z^2}=\dfrac{144}{23}=6\frac{6}{23}" align="absmiddle" class="latex-formula">

Question 3

Спасибо огромное)))

artalex74_zn · Answer 1 · 2018-03-12T10:24:33+0000

1.
$\dfrac{x^3+6x^2+4x-5}{x^2+8x+15}=1\\ \dfrac{(x^3+6x^2+4x-5)-(x^2+8x+15)}{(x+3)(x+5)}=0 \\ O.D.3.: x \neq-3,\ x \neq-5\\ x^3+5x^2-4x-20=0\\ x^2(x+5)-4(x+5)=0\\ (x+5)(x-2)(x+2)=0\\ x_1=-5,\ x_2=-2,\ x_3=2$
C учетом О.Д.З. получим х=-2, х=2.
Ответ: -2; 2.
2.
y=2z\\\\ \dfrac{x}{y}=3\ => x=3y=3(2z)=6z\\\\ \dfrac{4x^2+xy-2xz}{3y^2+2xz-z^2}=\dfrac{4(6z)^2+6z*2z-2z*6z}{3(2z)^2+2z*6z-z^2}=\\\\ =\dfrac{144z^2+12z^2-12z^2}{12z^2+12z^2-z^2}=\dfrac{144z^2}{23z^2}=\dfrac{144}{23}=6\frac{6}{23}" alt="\dfrac{y}{z}=2\ => y=2z\\\\ \dfrac{x}{y}=3\ => x=3y=3(2z)=6z\\\\ \dfrac{4x^2+xy-2xz}{3y^2+2xz-z^2}=\dfrac{4(6z)^2+6z*2z-2z*6z}{3(2z)^2+2z*6z-z^2}=\\\\ =\dfrac{144z^2+12z^2-12z^2}{12z^2+12z^2-z^2}=\dfrac{144z^2}{23z^2}=\dfrac{144}{23}=6\frac{6}{23}" align="absmiddle" class="latex-formula">