1.2-bo'lim: Kasr (ratsional) ko'rsatkichli darajalar (rational exponents)
Butun ko'rsatkichli darajalar bilan tanishib bo'lgach, endi ancha murakkabroq ko'rinishdagi darajalarni o'rganishga o'tamiz. Ushbu bo'limda biz ratsional (kasr) ko'rsatkichli darajalarni, ya'ni quyidagi ko'rinishdagi ifodalarni ko'rib chiqamiz:
\[{b^{\frac{m}{n}}}\]bu yerda \(m\) va \(n\) — butun sonlardir (\(n > 0\)).
Dastlab eng sodda maxsus holat — \({b^{\frac{1}{n}}}\) (bu yerda \(n\) — butun son) ko'rinishidan boshlaymiz. Ushbu holatni tushunib olsak, umumiy \(\frac{m}{n}\) holati o'z-o'zidan juda osonlashadi.
Avvalo, \({b^{\frac{1}{n}}}\) ifodaning ma'nosiga aniq ta'rif beraylik:
\[a = {b^{\frac{1}{n}}}\hspace{0.25in}\text{tenglik quyidagiga teng kuchlidir:}\hspace{0.25in}{a^n} = b\]Boshqacha aytganda, \({b^{\frac{1}{n}}}\) ni hisoblaganda biz: "Qaysi sonni (bu yerda \(a\) ni) \(n\)-darajaga ko'tarsak, natijada \(b\) hosil bo'ladi?" degan savolga javob izlaymiz. Odatda \({b^{\frac{1}{n}}}\) ifoda \(b\) sonining \(n\)-darajali ildizi deb ataladi.
- \({25^{\frac{1}{2}}}\)
- \({32^{\frac{1}{5}}}\)
- \({81^{\frac{1}{4}}}\)
- \({\left( { - 8} \right)^{\frac{1}{3}}}\)
- \({\left( { - 16} \right)^{\frac{1}{4}}}\)
- \( - {16^{\frac{1}{4}}}\)
a \({25^{\frac{1}{2}}}\) Yechim
Ushbu misolda biz: "Qaysi sonning kvadrati 25 ga teng?" deb so'rayapmiz:
\[{?^{\,2}} = 25 \implies 5^2 = 25\]Demak:
\[{25^{\frac{1}{2}}} = 5\]b \({32^{\frac{1}{5}}}\) Yechim
Qaysi sonning 5-darajasi 32 ga teng?
\[{32^{\frac{1}{5}}} = 2\hspace{0.25in}\text{chunki}\hspace{0.25in}{2^5} = 32\]c \({81^{\frac{1}{4}}}\) Yechim
Qaysi sonning 4-darajasi 81 ga teng?
\[{81^{\frac{1}{4}}} = 3\hspace{0.25in}\text{chunki}\hspace{0.25in}{3^4} = 81\]d \({\left( { - 8} \right)^{\frac{1}{3}}}\) Yechim
Minus ishorasiga e'tibor beramiz: qaysi sonning kubi \(-8\) ga teng?
\[{\left( { - 8} \right)^{\frac{1}{3}}} = - 2\hspace{0.25in}\text{chunki}\hspace{0.25in}{\left( { - 2} \right)^3} = - 8\]e \({\left( { - 16} \right)^{\frac{1}{4}}}\) Yechim
Ushbu misolning haqiqiy sonlar to'plamida yechimi mavjud emas! Chunki har qanday haqiqiy sonning juft (4-) darajasi musbat yoki nol bo'lishi shart, ya'ni hech qanday haqiqiy sonni 4-darajaga ko'tarib manfiy \(-16\) sonini hosil qilib bo'lmaydi.
E'tibor bering, bu oldingi (d) banddan farq qiladi: manfiy sonni toq darajaga ko'targanda manfiy natija chiqadi, ammo juft darajada haqiqiy yechim bo'lmaydi.
f \( - {16^{\frac{1}{4}}}\) Yechim
Bu misol qavslarning naqadar muhimligini yana bir bor ko'rsatadi. Bu yerda qavs yo'qligi sababli, \(\frac{1}{4}\) daraja faqat \(16\) soniga tegishli, minus ishorasi esa darajadan tashqarida qoladi:
\[ - {16^{\frac{1}{4}}} = - \left( {{{16}^{\frac{1}{4}}}} \right) = - \left( 2 \right) = - 2\]Umumiy ratsional daraja ko'rsatkichi: \({b^{\frac{m}{n}}}\)
Oldingida keltirilgan darajaning barcha 11 ta xossasi ixtiyoriy kasr ko'rsatkichli darajalar uchun ham to'liq o'rinlidir. Xususan, darajani darajaga ko'tarish xossasidan foydalanib (\({\left( {{a^n}} \right)^m} = {a^{nm}}\)), umumiy kasr darajani ikki xil teng kuchli usulda hisoblashimiz mumkin:
\[{b^{\frac{m}{n}}} = {\left( {{b^{\frac{1}{n}}}} \right)^m}\hspace{0.25in}\text{YOKI}\hspace{0.25in}{b^{\frac{m}{n}}} = {\left( {{b^m}} \right)^{\frac{1}{n}}}\]Amalda ko'pincha birinchi usul, ya'ni avval \(n\)-ildizni chiqarib, so'ngra \(m\)-darajaga ko'tarish ancha qulaydir (chunki kichikroq sonlar bilan ishlash osonroq bo'ladi).
- \({8^{\frac{2}{3}}}\)
- \({625^{\frac{3}{4}}}\)
- \({\left( {\displaystyle \frac{{243}}{{32}}} \right)^{\frac{4}{5}}}\)
a \({8^{\frac{2}{3}}}\) Yechim
Ikkala usulni ham ko'rib chiqaylik:
1-usul: \({\left( {{8^{\frac{1}{3}}}} \right)^2} = {\left( 2 \right)^2} = 4\)
2-usul: \({\left( {{8^2}} \right)^{\frac{1}{3}}} = {\left( {64} \right)^{\frac{1}{3}}} = 4\)
Har ikkala usulda ham natija bir xil chiqadi, lekin birinchi usulda sonlar kichikroq va hisoblash ancha tez kechadi.
b \({625^{\frac{3}{4}}}\) Yechim
Birinchi usul bilan hisoblaymiz:
\[{625^{\frac{3}{4}}} = {\left( {{{625}^{\frac{1}{4}}}} \right)^3} = {\left( 5 \right)^3} = 125\](Agar 2-usul qo'llanilsa, avval \(625^3 = 244140625\) ni hisoblashga to'g'ri kelardi, bu esa hisoblashni o'ta qiyinlashtiradi).
c \({\left( {\displaystyle \frac{{243}}{{32}}} \right)^{\frac{4}{5}}}\) Yechim
Kasr darajasini surat va maxrajga alohida taqsimlaymiz:
\[{\left( {\frac{{243}}{{32}}} \right)^{\frac{4}{5}}} = \frac{{{{243}^{\frac{4}{5}}}}}{{{{32}^{\frac{4}{5}}}}} = \frac{{{{\left( {{{243}^{\frac{1}{5}}}} \right)}^4}}}{{{{\left( {{{32}^{\frac{1}{5}}}} \right)}^4}}} = \frac{{{{\left( 3 \right)}^4}}}{{{{\left( 2 \right)}^4}}} = \frac{{81}}{{16}}\]- \({\left( {\displaystyle \frac{{{w^{ - 2}}}}{{16{v^{\frac{1}{2}}}}}} \right)^{\frac{1}{4}}}\)
- \({\left( {\displaystyle \frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}}\)
a \({\left( {\displaystyle \frac{{{w^{ - 2}}}}{{16{v^{\frac{1}{2}}}}}} \right)^{\frac{1}{4}}}\) Yechim
Tashqaridagi \(\frac{1}{4}\) darajani har bir ko'paytuvchiga kiritamiz:
\[\frac{{{w^{ - 2\left( {\frac{1}{4}} \right)}}}}{{{{16}^{\frac{1}{4}}}{v^{\frac{1}{2}\left( {\frac{1}{4}} \right)}}}} = \frac{{{w^{ - \frac{1}{2}}}}}{{2{v^{\frac{1}{8}}}}}\]Manfiy darajali \(w^{-\frac{1}{2}}\) ni maxrajga tushirib, musbat darajaga aylantiramiz:
\[= \frac{1}{{2{v^{\frac{1}{8}}}{w^{\frac{1}{2}}}}}\]b \({\left( {\displaystyle \frac{{{x^2}{y^{ - \frac{2}{3}}}}}{{{x^{ - \frac{1}{2}}}{y^{ - 3}}}}} \right)^{ - \frac{1}{7}}}\) Yechim
Dastlab qavs ichidagi kasrni soddalashtirib olamiz:
\[{\left( {\frac{{{x^2}{x^{\frac{1}{2}}}{y^3}}}{{{y^{\frac{2}{3}}}}}} \right)^{ - \frac{1}{7}}} = {\left( {{x^{2 + \frac{1}{2}}}{y^{3 - \frac{2}{3}}}} \right)^{ - \frac{1}{7}}} = {\left( {{x^{\frac{5}{2}}}{y^{\frac{7}{3}}}} \right)^{ - \frac{1}{7}}}\]Tashqaridagi manfiy darajadan qutulish uchun ifodani maxrajga tushiramiz va darajani ko'paytiramiz:
\[= \frac{1}{{{{\left( {{x^{\frac{5}{2}}}{y^{\frac{7}{3}}}} \right)}^{\frac{1}{7}}}}} = \frac{1}{{{x^{\frac{5}{{14}}}}{y^{\frac{1}{3}}}}}\]Ushbu bo'limni yakunlashda juda ko'p uchraydigan muhim bir xatodan ogohlantirib o'tmoqchimiz: manfiy daraja va kasr darajani hech qachon bir-biri bilan adashtirmang!
\[{b^{ - n}} = \frac{1}{{{b^n}}}\hspace{0.5in}\text{lekin}\hspace{0.5in}{b^{ - n}} \ne {b^{\frac{1}{n}}}\]