2.9-bo'lim : Shaklda kvadratga tushiriladigan tenglamalar
5. Quyidagi tenglamani yeching.
\[\frac{2}{{{x^2}}} + \frac{{17}}{x} + 21 = 0\]
Maslahat: Bu birinchi to'rtta muammo bilan bir xil ishlaydi, garchi \(x\) maxrajda bo'lsa ham. Bu erdagi yagona farq shundaki, \(x\)lar bizning almashtirish maxrajida bo'ladi.
Yechim
Birinchidan, \(2 = 2\left( 1 \right)\) va shuning uchun tenglamani kvadrat tenglamaga kamaytirish uchun quyidagi almashtirishdan foydalanishimiz mumkinligiga e'tibor bering.
\[u = \frac{1}{x}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {\frac{1}{x}} \right)^2} = \frac{{{1^2}}}{{{x^2}}} = \frac{1}{{{x^2}}}\]2-qadam
Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}2{u^2} + 17u + 21 & = 0\\ \left( {2u + 3} \right)\left( {u + 7} \right) & = 0\end{align*}\]Bu tenglamaning yechimi \(u = - \frac{3}{2}\) va \(u = - 7\) ekanligini osongina ko'rishimiz mumkin.
3-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa, dastlabki tenglamaning echimini aniqlash uchun birinchi bosqichdan boshlab almashtirishimizdan foydalanishdir.
\[u = - \frac{3}{2}:\hspace{0.25in}\frac{1}{x} = - \frac{3}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{{ - \frac{3}{2}}} = - \frac{2}{3}\] \[u = - 7:\hspace{0.25in}\frac{1}{x} = - 7\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{{ - 7}} = - \frac{1}{7}\]Shuning uchun dastlabki tenglamaning ikkita yechimi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{2}{3}\,\,{\mbox{and }}x = - \frac{1}{7}}}\].