Question

Помогите пожалуйста алгебра номер 8.22.

Answer 1

$3x^2+8x-1=0\; \; \; \to \; \; \; \; \left \{ {{x_1\cdot x_2=-\frac{1}{3}} \atop {x_1+x_2=-\frac{8}{3}}} \right.\\\\\\1)\; \; (x_1+x_2)^2=x^2_1+x_2^2+2x_1x_2\; \; ,\; \; (-\frac{8}{3})^2=x_1^2+x_2^2+2\cdot (-\frac{1}{3})\\\\x_1^2+x_2^2=\frac{64}{9}+2x_2x_2=\frac{64}{9}+\frac{2}{3}=\frac{70}{9}$

$2)\; \; x_1x_2^3+x_2x_1^3=x_1x_2\cdot (x_2^2+x_1^2)=-\frac{1}{3}\cdot \frac{70}{9}=-\frac{70}{27}$

$3)\; \; (x_1+x_2)^3=x_1^3+x_2^3+3x_1x_2\cdot (x_1+x_2)\; \; \to \; \; (-\frac{8}{3})^3=x_1^3+x_2^3+3\cdot (-\frac{1}{3})\cdot (-\frac{8}{3})\\\\-\frac{512}{27}=x_1^3+x_2^3+\frac{8}{3}\; \; \to \; \; \; x_1^3+x_2^3=-\frac{512}{27}-\frac{8}{3}=-\frac{584}{27}\\\\\frac{x_1}{x_2^2}+\frac{x_2}{x_1^2}=\frac{x_1^3+x_2^3}{x_1^2x_2^2}=\frac{-584}{27\cdot (-\frac{1}{3})^2}=-\frac{584}{3}$

$4)\; \; (x_1^2+x_2^2)^2=x_1^4+x_2^4+2x_1^2x_2^2\; \; ,\; \; (\frac{70}{9})^2=x_1^4+x_2^4+2\cdot (-\frac{1}{3})^2\\\\x_1^4+x_2^4=\frac{4900}{81}-\frac{2}{9}=\frac{4882}{81}$