7.5-bo'lim : Nochiziqli tizimlar
3. Quyidagi tenglamalar sistemasining yechimini toping.
\[\begin{align*}xy & = 4\\ \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} & = 1\end{align*}\]Yechim
Yechimga juda uzoqqa kirishdan oldin, birinchi navbatda, bularni hal qilishning yagona to'g'ri yo'li yo'qligini ta'kidlashimiz kerak. Ushbu turdagi muammolarning ko'pchiligi yechim topish uchun bir nechta yo'llarga ega bo'ladi. Biroq, qanday yo'l tutishimizdan qat'i nazar, tizimga yechim bir xil bo'ladi.
OK, muammoga o'tamiz. Bu holda biz \(x\) yoki \(y\) uchun birinchi tenglamani yechib, ikkinchi tenglamaga qo'shamiz. Hech qanday sababsiz bizda \(x\)-da ushbu bo'lim uchun dastlabki ikkita mashq muammosi uchun tenglamalar mavjud edi, biz \(x\) uchun birinchi tenglamani yechamiz va uni ikkinchi tenglamaga qo'shamiz. Natijada faqat \(y\) ni o'z ichiga olgan tenglama bo'ladi.
Mana o'sha ish.
\[\begin{align*}x = \frac{4}{y}\hspace{0.25in} \to \hspace{0.25in}\frac{{{{\left( {\frac{4}{y}} \right)}^2}}}{4} + \frac{{{y^2}}}{{25}} & = 1\\ \hspace{0.25in}\frac{{\frac{{16}}{{{y^2}}}}}{4} + \frac{{{y^2}}}{{25}} & = 1\\ \hspace{0.25in}\frac{4}{{{y^2}}} + \frac{{{y^2}}}{{25}} & = 1\end{align*}\]2-qadam
Keling, maxrajlarni tozalash uchun ikkala tomonini \(25{y^2}\) ga ko'paytiramiz.
\[\begin{align*}\,100 + {y^4} & = 25{y^2}\\ {y^4} - 25{y^2} + 100 & = 0\end{align*}\]3-qadam
Bu kvadratik shaklga ega, shuning uchun biz \(u = {y^2}\) (va shuning uchun \({u^2} = {\left( {{y^2}} \right)^2} = {y^4}\)) ni aniqlashimiz mumkin. Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}{u^2} - 25u + 100 & = 0\\ \left( {u - 5} \right)\left( {u - 20} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}u = 5,\,\,\,\,\,\,\,u = 20\end{align*}\]4-qadam
Shunday qilib, biz \(u\) ning ikkita qiymatini oldik va ularning har biri \(y\) nuqtai nazaridan quyidagi tenglamaga mos keladi (i.e. yuqoridagi almashtirish yordamida).
\[\begin{align*} & u = 5\,\,\,\,:\,\,\,\,\,\,{y^2} = 5 & \to \hspace{0.25in} & y = \pm \sqrt 5 \\ & u = 20:\,\,\,\,\,\,{y^2} = 20 & \to \hspace{0.25in} & y = \pm \sqrt {20} = \pm 2\sqrt 5 \end{align*}\]
5-qadam
Bizda \(y\) ning to'rtta qiymati bor, ular uchun \(x\) ning mos qiymatlarini topishimiz kerak. Biz ularni birinchi tenglamaga qo'shamiz (bu tenglamani qo'shish ancha oson).
\[\begin{align*} & y = \sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( {\frac{4}{{\sqrt 5 }},\sqrt 5 } \right)\\ & y = - \sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{ - \sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( { - \frac{4}{{\sqrt 5 }}, - \sqrt 5 } \right)\\ & y = 2\sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{2\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( {\frac{2}{{\sqrt 5 }},2\sqrt 5 } \right)\\ & y = - 2\sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{ - 2\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( { - \frac{2}{{\sqrt 5 }}, - 2\sqrt 5 } \right)\end{align*}\]Shunday qilib, ushbu tenglamalar sistemasi uchun yuqorida sanab o'tilgan to'rtta yechim mavjud.