Question

Решите систему уравнений способом сложения х^2 +у^2=81 х^2-у=9 помогите срочно пжж​

Answer 1

$\begin{cases}x^2+y^2=81\\x^2-y=9\quad|\cdot(-1)\end{cases}\Longleftrightarrow\quad+\begin{cases}x^2+y^2=81\\-x^2+y=-9\end{cases}\Longleftrightarrow\quad\begin{cases}y^2+y=72\\x^2-y=9\end{cases}$

$y^2+y=72\\\\y^2+y-72=0$

$D=1^2-4\cdot1(-72)=1+288=289 \qquad\qquad\boxed{D = b^2-4ac}$

289 > 0 ⇒  2 корня

$\sqrt{289}=17\\\\\left\begin{array}{lcl}y_1=\dfrac{-1+17}{2}=8\\\\y_2=\dfrac{-1-17}{2}=-9\end{array}\right \qquad\qquad\boxed{y=\dfrac{-b\underline+\sqrt{D}}{2a}}$

$x^2-8=9\\x^2=17\\\\x=\sqrt{17}\\x=-\sqrt{17}$

$x^2-(-9)=9\\x^2=0\\\\x=0$

$x_1=-\sqrt{17};\ x_2=0;\ x_3=\sqrt{17}$

Ответ:  $(-\sqrt17;\ 8);\ (0;\ -9);\ (\sqrt{17};\ 8)$

Answer 2

$\displaystyle\bf\\-\left \{ {{x^{2} +y^{2}=81 } \atop {x^{2} -y =9}} \right. \\---------\\y^{2} +y=72\\\\y^{2} +y-72=0\\\\Teorema \ Vieta:\\\\y_{1} =-9 \ ; \ y_{2} =8\\\\x^{2} =9+y_{1} =9-9=0\\\\x_{1} =0\\\\\\x^{2} =9+y_{2}=9+8= 17\\\\x_{2,3} =\pm\sqrt{17} \\\\Otvet:\boxed{(0 \ ; \ -9) \ , \ (-\sqrt{17} \ ; \ 8) \ , \ (\sqrt{17} \ ; \ 8)}$