Question

Помогите с 1,2,3,4 пожалуйста

Answer 1

Ответ:

$1)\ \ x^2+3x-40=(x-5)(x+8)\ \ ,\ \ \ \ 6x^2+x-12=(2x+3)(3x-4)\\\\\\2)\ a)\ \ x^4-15x^2-16=0\ \ ,\ \ x^2_1=16\ \ ,\ \ x^2_2=-1<0\ \ (teorema\ Vieta)\\\\x^2=16\ \ \to \ \ x=\pm 4\\\\b)\ \ \dfrac{x^2+12}{x-3}=\dfrac{7x}{x-3}\ \ \ ,\ \ \ \ ODZ:\ x\ne 3\\\\x^2+12=7x\ \ ,\ \ \ x^2-7x+12=0\ \ ,\ \ x_1=3\notin ODZ\ ,\ \ x_2=4\\\\Otvet:\ \ x=4\ .\\\\\\3)\ \ \dfrac{5a^2+3a-2}{a^2-1}=\dfrac{(a+1)(5a-2)}{(a-1)(a+1)}=\dfrac{5a-2}{a-1}$

$4)\ \ \dfrac{4}{x^2-10x+25}-\dfrac{10}{x^2-25}=\dfrac{1}{x+5}\ \ ,\ \ \ ODZ:\ x\ne \pm 5\\\\\\\dfrac{4}{(x-5)^2}-\dfrac{10}{(x-5)(x+5)}=\dfrac{1}{x+5}\\\\\\\dfrac{4(x+5)-10(x-5)-(x-5)^2}{(x-5)^2(x+5)}=0\\\\\\\dfrac{-(x^2-4x-45)}{(x-5)^2(x+5)}=0\ \ ,\ \ \ \dfrac{(x+9)(x-5)}{(x-5)^2(x+5)}=0\ \ \to \ \ \ x_1=-9\ ,\ x_2=5\\\\\\Otvet:\ \ x=-9\ .$