Question

Срочно, иначе я умруууууууу!!!

Answer 1

Ответ:

$ax^2+bx+c=0\ \ \ \Rightarrow \ \ \ \left\{\begin{array}{l}x_1\cdot x_2=\dfrac{c}{a}\\x_1+x_2=-\dfrac{b}{a}\end{array}\right\ \ \ teorema\ Vieta\\\\\\a)\ \ x_1'=x_1+\dfrac{1}{x_1}\ \ ,\ \ \ x_2'=x_1+\dfrac{1}{x_1}\ \ ,\\\\\\x_1'\cdot x_2'=\bigg(x_1+\dfrac{1}{x_1}\bigg)\bigg(x_2+\dfrac{1}{x_2}\bigg)=\dfrac{x_1^2+1}{x_1}\cdot \dfrac{x_2^2+1}{x_2}=\dfrac{(x_1x_2)^2+x_1^2+x_2^2+1}{x_1\cdot x_2}=$

$=\dfrac{\frac{c^2}{a^2}+(x_1+x_2)^2-2x_1x_2+1}{\frac{c}{a}}=\dfrac{\frac{c^2}{a^2}+\frac{b^2}{a^2}-2\cdot \frac{c}{a}+1}{\frac{c}{a}}=\dfrac{c^2+b^2-2ac+a^2}{ac}=\\\\\\=\dfrac{(a-c)^2+b^2}{ac}$

$x_1'+x_2'=\bigg(x_1+\dfrac{1}{x_1}\bigg)+\bigg(x_2+\dfrac{1}{x_2}\bigg)=\dfrac{x_1^2+1}{x_1}+\dfrac{x_2^2+1}{x_2}=\dfrac{x_2(x_1^2+1)+x_1(x_2^2+1)}{x_1\cdot x_2}=\\\\\\=\dfrac{x_1^2x_2+x_2+x_1x_2^2+x_1}{x_1\cdot x_2}=\dfrac{x_1x_2(x_1+x_2)+(x_1+x_2)}{x_1x_2}=\dfrac{\frac{c}{a}\cdot (-\frac{b}{a})-\frac{b}{a}}{\frac{c}{a}}=\\\\\\=\dfrac{-\frac{bc}{a^2}-\frac{b}{a}}{\frac{c}{a}}=\dfrac{-bc-ab}{ac}=-\dfrac{b(a+c)}{ac}\\\\\\x^2+\dfrac{b(a+c)}{ac}\, x+\dfrac{(a-c)^2+b^2}{ac}=0$

$\underline {\ ac\, x^2+b(a+c)\, x+(a-c)^2+b^2=0\ }$

$b)\ \ x_1'=\dfrac{x_1+1}{x_1-1}\ \ ,\ \ x_2'=\dfrac{x_2+1}{x_2-1}\\\\\\x_1'\cdot x_2'=\dfrac{x_1+1}{x_1-1}\cdot \dfrac{x_2+1}{x_2-1}=\dfrac{x_1x_2+x_1+x_2+1}{x_1x_2-x_1-x_2+1}=\dfrac{x_1x_2+(x_1+x_2)+1}{x_1x_2-(x_1+x_2)+1}=\\\\\\=\dfrac{\frac{c}{a}-\frac{b}{a}+1}{\frac{c}{a}+\frac{b}{a}+1}=\dfrac{c-b+a}{c+b+a}=\dfrac{a+c-b}{a+b+c}$

$x_1'+x_2'=\dfrac{x_1+1}{x_1-1}+\dfrac{x_2+1}{x_2-1}=\dfrac{(x_1+1)(x_2-1)+(x_2+1)(x_1-1)}{(x_1-1)(x_2-1)}=\\\\\\=\dfrac{x_1x_2-x_1+x_2-1+x_1x_2-x_2+x_1-1}{x_1x_2-x_1-x_2+1}=\dfrac{2x_1x_2-2}{x_1x_2-(x_1+x_2)+1}=\\\\\\=\dfrac{\frac{2c}{a}-2}{\frac{c}{a}+\frac{b}{a}+1}=\dfrac{2c-2a}{c+b+a}=\dfrac{2(c-a)}{b+c+a}=\dfrac{2(c-a)}{a+b+c}\\\\\\\\x^2-\dfrac{2(c-a)}{a+b+c}\, x+\dfrac{a+c-b}{a+b+c}=0\\\\\\\underline{\ (a+b+c)\, x^2-2(c-a)\, x+(a+c-b)=0\ }$