Question

Помогите Пожалуйста,на фото отмечено что делать

Answer 1

Ответ:

$5.1.d)\ \ \ 3x^2-8y\cdot x+5y^2=0\\\\D/4=(\frac{b}{2})^2-ac=(-4y)^2-3\cdot 4y^2=18y^2-15y^2=y^2\\\\x_{1,2}=\dfrac{-\frac{b}{2}-\sqrt{D/4}}{a}=\dfrac{4y\pm y}{3}\\\\x_1=\dfrac{4y-y}{3}=y\ \ ,\ \ \ x_1=\dfrac{4y+y}{3}=\dfrac{5}{3}\, y$

$5.4.b)\ \ \ \sigma _1=x+y\ \ ,\ \ \sigma _2=xy\\\\S_1=x+y=\sigma _1\\\\S_2=x^2+y^2=\sigma _1^2-2\sigma _2\ \ ,\ \ \ S_3=x^3+y^3=\sigma _1^3-3\sigma _1\sigma _2\ \ ,\\\\\\x^3-2x^2y^2+y^3=(x^3+y^3)-2\cdot (xy)^2=S_3-2\sigma _2^2=\sigma _1^3-3\sigma _1\sigma _2-2\sigma _2^2$

$5.10)\ \ \ 4x^2-6x-1=0\ \ ,\ \ \left\{\begin{array}{l}\alpha +\beta =\dfrac{6}{4}=\dfrac{3}{2}\\\alpha \cdot \beta =-\dfrac{1}{4}\end{array}\right\ \ \ (teorema\ Vieta)$

$\left\{\begin{array}{l}\alpha =\dfrac{3}{2}-\beta \\(\dfrac{3}{2}-\beta )\cdot \beta =-\dfrac{1}{4}\end{array}\right\ \ \left\{\begin{array}{l}\alpha =\dfrac{3}{2}-\beta \\\beta ^2-\dfrac{3}{2}\beta -\dfrac{1}{4}=0\end{array}\right\ \ \left\{\begin{array}{l}\alpha =\dfrac{3}{2}-\beta \\\beta _{1}=\dfrac{3-\sqrt{13}}{4}\ ,\ \beta_2=\dfrac{3+\sqrt{13}}{4} \end{array}\right$

$\left\{\begin{array}{l}\alpha _1=\dfrac{3+\sqrt{13}}{4}\ ,\ \alpha _2=\dfrac{3-\sqrt{13}}{4}\\\beta _{1}=\dfrac{3-\sqrt{13}}{4}\ ,\ \beta _2=\dfrac{3+\sqrt{13}}{4}\end{array}\right\\\\\\a)\ \ x_1=\dfrac{\alpha _1}{\beta _1}+1\ \ ,\ \ \ x_2=\dfrac{\beta _1}{\alpha _1}+1\\\\\\x_1=\dfrac{3+\sqrt{13}}{3-\sqrt{13}}+1=\dfrac{(3+\sqrt{13})^2}{9-13}+1=\dfrac{22+6\sqrt{13}}{9-13}+1=\dfrac{-9-3\sqrt{13}}{2}$

$x_2=\dfrac{3-\sqrt{13}}{3+\sqrt{13}}+1=\dfrac{(3-\sqrt{13})^2}{9-13}+1=\dfrac{22-6\sqrt{13}-4}{-4}=\dfrac{-9+3\sqrt{13}}{2}$

$x^2+px+q=0\ \ \ \Rightarrow \\\\-p=x_1+x_2=\dfrac{-9-3\sqrt{13}}{2}+\dfrac{-9+3\sqrt{13}}{2}=-\dfrac{18}{2}=-9\ \ ,\ \ \ p=9\ \ ,\\\\q=x_1\cdot x_2=\dfrac{-(9+3\sqrt{13})}{2}\cdot \dfrac{-9+3\sqrt{13}}{2}=\dfrac{-(9\cdot 13-81)}{4}=-9\\\\\underline {\ x^2+9\, x-9=0\ }$

$b)\ \ x_1=\dfrac{\alpha _2}{\beta _2}+1\ \ ,\ \ \ x_2=\dfrac{\beta _2}{\alpha _2}+1\\\\\\x_1=\dfrac{-9+3\sqrt{13}}{2}\ \ ,\ \ x_2=\dfrac{-9-3\sqrt{13}}{2}\\\\\\x^2+px+q=0\ \ \ ,\ \ \ \ \underline {\ x^2+9\, x-9=0\ }$