2.9-bo'lim : Shaklda kvadratga tushiriladigan tenglamalar
1. Quyidagi tenglamani yeching.
\[{x^6} - 9{x^3} + 8 = 0\]
Maslahat: Birinchi ikkita hadning ko'rsatkichlariga qarashni unutmang va uni "normal" kvadrat tenglamaga aylantiradigan almashtirishni topishga harakat qiling.
Yechim
Birinchidan, \(6 = 2\left( 3 \right)\) va shuning uchun tenglamani kvadrat tenglamaga kamaytirish uchun quyidagi almashtirishdan foydalanishimiz mumkinligiga e'tibor bering.
\[u = {x^3}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^3}} \right)^2} = {x^6}\]2-qadam
Ushbu almashtirishdan foydalanib, tenglama quyidagicha bo'ladi:
\[\begin{align*}{u^2} - 9u + 8 & = 0\\ \left( {u - 1} \right)\left( {u - 8} \right) & = 0\end{align*}\]Bu tenglamaning yechimi \(u = 1\) va \(u = 8\) 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.
\[u = 1:\hspace{0.25in}{x^3} = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( 1 \right)^{\frac{1}{3}}} = 1\] \[u = 8:\hspace{0.25in}{x^3} = 8\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( 8 \right)^{\frac{1}{3}}} = 2\]Shuning uchun asl tenglamaning ikkita yechimi: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 1\,\,{\mbox{and }}x = 2}}\).