2.9-bo'lim : Shaklda kvadratga tushiriladigan tenglamalar
3. Quyidagi tenglamani yeching.
\[4{x^{\frac{2}{3}}} + 21{x^{\frac{1}{3}}} + 27 = 0\]
Maslahat: Birinchi ikkita hadning ko'rsatkichlariga qarashni unutmang va uni "normal" kvadrat tenglamaga aylantiradigan almashtirishni topishga harakat qiling.
Yechim
Birinchidan, \(\frac{2}{3} = 2\left( {\frac{1}{3}} \right)\) va shuning uchun tenglamani kvadrat tenglamaga kamaytirish uchun quyidagi almashtirishdan foydalanishimiz mumkinligiga e'tibor bering.
\[u = {x^{\frac{1}{3}}}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^{\frac{1}{3}}}} \right)^2} = {x^{\frac{2}{3}}}\]2-qadam
Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}4{u^2} + 21u + 27 & = 0\\ \left( {4u + 9} \right)\left( {u + 3} \right) & = 0\end{align*}\]Bu tenglamaning yechimi \(u = - \frac{9}{4}\) va \(u = - 3\) 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.
\[u = - \frac{9}{4}:\hspace{0.25in}{x^{\frac{1}{3}}} = - \frac{9}{4}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( { - \frac{9}{4}} \right)^3} = - \frac{{729}}{{64}}\] \[u = - 3:\hspace{0.25in}{x^{\frac{1}{3}}} = - 3\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( { - 3} \right)^3} = - 27\]Shuning uchun dastlabki tenglamaning ikkita yechimi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{729}}{{64}}\,\,{\mbox{and }}x = - 27}}\].