Question

алгебра 8 класс. СРОЧНО!!!​

Answer 1

$x_{1}+x_{2}=-frac{b}{a}$

$x_{1}cdot x_{2}=frac{c}{a}$

1.

$frac{1}{x_1^2}+ frac{1}{x_2^2}=frac{x_2^2+x_1^2}{x_1^2x_2^2}=frac{x_2^2+x_1^2}{(x_1x_2)^2}= frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2} =frac{frac{b^2-2ac}{a^2}}{left(frac{c}{a}right)^2}= frac{b^2-2ac}{c^2}$

$2x^2-5x-4=0$

$a=2, b=-5, c=-4$

$frac{1}{x_1^2}+ frac{1}{x_2^2}=frac{b^2-2ac}{c^2}= frac{(-5)^2-2 cdot 2 cdot (-4)}{(-4)^2} =frac{25+16}{16} =frac{41}{16}$

2.

$x_1x_2^4+x_2x_1^4=x_1x_2(x_2^3+x_1^3)=]x_1x_2(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=$

$x_1x_2(x_1+x_2)((x_1+x_2)^2-2x_1x_2-x_1x_2)=x_1x_2(x_1+x_2)left[(x_1+x_2)^2-3x_1x_2right] =$

$frac{c}{a}cdotleft(-frac{b}{a}right) cdotleft[left(-frac{b}{a}right) ^2-3cdotfrac{c}{a}right] -frac{bc}{a^2}cdotleft(frac{b^2}{a^2}-frac{3c}{a}lright) =$

$-frac{bc}{a^2}cdotfrac{b^2-3ac}{a^2}=-frac{bc(b^2-3ac)}{a^4}$

$2x^2-5x-4=0$

$a=2, b=-5, c=-4$

$x_1x_2^4+x_2x_1^4=-frac{bc(b^2-3ac)}{a^4}=-frac{-5 cdot (-4)cdot((-5)^2-3 cdot 2 cdot (-4))}{2^4}=-frac{20cdot(25+24}{16}=-frac{5cdot49}{4}=-frac{245}{4}$