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.

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

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

$\frac{c}{a}\cdot\left(-\frac{b}{a}\right) \cdot\left[\left(-\frac{b}{a}\right) ^2-3\cdot\frac{c}{a}\right] -\frac{bc}{a^2}\cdot\left(\frac{b^2}{a^2}-\frac{3c}{a}\l\right) =$

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

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

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

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