2.9-bo'lim : Shaklda kvadratga tushiriladigan tenglamalar
6. Quyidagi tenglamani yeching.
\[\frac{1}{x} - \frac{{11}}{{\sqrt x }} + 18 = 0\]
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
Avval \(1 = 2\left( {\frac{1}{2}} \right)\) ga e'tibor beraylik va \(\sqrt x = {x^{\frac{1}{2}}}\) ni eslaylik. Shunday qilib, biz tenglamani kvadrat tenglamaga kamaytirish uchun quyidagi almashtirishdan foydalanishimiz mumkin.
\[u = \frac{1}{{\sqrt x }}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {\frac{1}{{\sqrt x }}} \right)^2} = {\left( {\frac{1}{{{x^{\frac{1}{2}}}}}} \right)^2} = \frac{1}{x}\]2-qadam
Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}{u^2} - 11u + 18 & = 0\\ \left( {u - 2} \right)\left( {u - 9} \right) & = 0\end{align*}\]Bu tenglamaning yechimi \(u = 2\) va \(u = 9\) 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.
\[u = 2:\hspace{0.25in}\frac{1}{{\sqrt x }} = 2\hspace{0.25in} \Rightarrow \hspace{0.25in}\sqrt x = \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\] \[u = 9:\hspace{0.25in}\frac{1}{{\sqrt x }} = 9\hspace{0.25in} \Rightarrow \hspace{0.25in}\sqrt x = \frac{1}{9}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {\frac{1}{9}} \right)^2} = \frac{1}{{81}}\]Shuning uchun dastlabki tenglamaning ikkita yechimi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{1}{4}\,\,{\mbox{and }}x = \frac{1}{{81}}}}\].