1.1-bo'lim: Butun ko'rsatkichli darajalar (integer exponents)
Ushbu bobni butun ko'rsatkichli darajalarni o'rganishdan boshlaymiz. Dastlab daraja ko'rsatkichlari musbat butun sonlar deb faraz qilamiz, so'ngra nol va manfiy darajalarni ham batafsil ko'rib chiqamiz.
Avvalo, musbat butun ko'rsatkichli darajaga ko'tarish ta'rifini eslaylik. Agar \(a\) — ixtiyoriy son va \(n\) — musbat butun son bo'lsa, u holda:
\[{a^n} = \underbrace {a \cdot a \cdot a \cdot \cdots \cdot a}_{n\text{ marta}}\]Masalan:
\[{3^5} = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 243\]Shuningdek, ushbu o'rinda darajaga ko'tarishda qavslarning o'rni va qoidalarini alohida eslatib o'tish lozim. Bu, ayniqsa, manfiy sonlar bilan ishlaganda o'ta muhimdir. Quyidagi ikki holatni solishtiring:
\[{\left( { - 2} \right)^4}\hspace{0.5in}\text{va}\hspace{0.5in} - {2^4}\]Ushbu ifodalarning hisoblash natijalari butunlay boshqacha bo'ladi! Darajaga ko'tarish amalida qat'iy qoida shundaki: daraja ko'rsatkichi faqat uning bevosita chap tomonida turgan ifodagagina taalluqlidir!
Birinchi holatda daraja ko'rsatkichining bevosita chap tomonida qavs turibdi, demak qavs ichidagi hamma narsa 4-darajaga ko'tariladi:
\[{\left( { - 2} \right)^4} = \left( { - 2} \right)\left( { - 2} \right)\left( { - 2} \right)\left( { - 2} \right) = 16\]Ikkinchi holatda esa daraja ko'rsatkichining bevosita chap tomonida faqat \(2\) turibdi. Shuning uchun faqat \(2\) soni darajaga ko'tariladi, oldidagi minus ishorasi esa darajaga ko'tarilmay, tashqarida qoladi:
\[ - {2^4} = - \left( {{2^4}} \right) = - \left( {2 \cdot 2 \cdot 2 \cdot 2} \right) = - 16\]Bu tushuntirishdan ko'zlangan asosiy maqsad — qavslarga nihoyatda e'tiborli bo'lishingizni ta'minlashdir. Qavslarni e'tiborsiz qoldirish yoki noo'rin joyga qavs qo'yish masalaning javobini mutlaqo o'zgartirib yuborishi mumkin. Bu ogohlantirish nafaqat darajalar uchun, balki butun matematika kursi davomida amal qiladi.
Nol va manfiy daraja ko'rsatkichlari
Nol daraja uchun quyidagi qoida o'rinli:
\[{a^0} = 1\hspace{0.5in}\text{(\(a \ne 0\) shartida)}\]E'tibor bering, \(a \ne 0\) bo'lishi shart, chunki \(0^0\) ifoda matematikada aniqlanmagan (mavjud emas). Masalan:
\[{\left( { - 1268} \right)^0} = 1\]Manfiy butun daraja ko'rsatkichi uchun quyidagi ta'rif qabul qilingan: agar \(a\) — noldan farqli ixtiyoriy son va \(n\) — musbat butun son bo'lsa, u holda:
\[{a^{ - n}} = \frac{1}{{{a^n}}}\]Nega \(a \ne 0\) talab qilinishini tushundingizmi? Chunki nolga bo'lish mumkin emas; agar \(a = 0\) bo'lishiga ruxsat berilsa, maxrajda nol hosil bo'lar edi. Quyida bir nechta misollar keltirilgan:
\[{5^{ - 2}} = \frac{1}{{{5^2}}} = \frac{1}{{25}}\hspace{0.5in}\hspace{0.25in}{\left( { - 4} \right)^{ - 3}} = \frac{1}{{{{\left( { - 4} \right)}^3}}} = \frac{1}{{ - 64}} = - \frac{1}{{64}}\]Darajaning asosiy xossalari (Properties of Exponents)
| 1. \({a^n}{a^m} = {a^{n + m}}\) | Misol : \({a^{ - 9}}{a^4} = {a^{ - 9 + 4}} = {a^{ - 5}}\) |
| 2. \({\left( {{a^n}} \right)^m} = {a^{nm}}\) | Misol : \({\left( {{a^7}} \right)^3} = {a^{\left( 7 \right)\left( 3 \right)}} = {a^{21}}\) |
| 3. \(\displaystyle \frac{{{a^n}}}{{{a^m}}} = \left\{ {\begin{array}{l}{{a^{n - m}}}\\{\displaystyle \frac{1}{{{a^{m - n}}}}}\end{array}\,\,\,,\,\,\,\,a \ne 0} \right.\) | Misol : \(\begin{align*}\frac{{{a^4}}}{{{a^{11}}}} & = {a^{4 - 11}} = {a^{ - 7}}\\ \frac{{{a^4}}}{{{a^{11}}}} & = \frac{1}{{{a^{11 - 4}}}} = \frac{1}{{{a^7}}} = {a^{ - 7}}\end{align*}\) |
| 4. \({\left( {ab} \right)^n} = {a^n}{b^n}\) | Misol : \({\left( {ab} \right)^{ - 4}} = {a^{ - 4}}{b^{ - 4}}\) |
| 5. \({\displaystyle \left( {\frac{a}{b}} \right)^n} = \displaystyle \frac{{{a^n}}}{{{b^n}}},\,\,\,\,b \ne 0\) | Misol : \({\left( {\displaystyle \frac{a}{b}} \right)^8} = \displaystyle \frac{{{a^8}}}{{{b^8}}}\) |
| 6. \(\displaystyle {\left( {\frac{a}{b}} \right)^{ - n}} = {\left( {\frac{b}{a}} \right)^n} = \frac{{{b^n}}}{{{a^n}}}\) | Misol : \({\left( {\displaystyle \frac{a}{b}} \right)^{ - 10}} = {\left( {\displaystyle \frac{b}{a}} \right)^{10}} = \displaystyle \frac{{{b^{10}}}}{{{a^{10}}}}\) |
| 7. \({\left( {ab} \right)^{ - n}} = \displaystyle \frac{1}{{{{\left( {ab} \right)}^n}}}\) | Misol : \({\left( {ab} \right)^{ - 20}} = \displaystyle \frac{1}{{{{\left( {ab} \right)}^{20}}}}\) |
| 8. \(\displaystyle \frac{1}{{{a^{ - n}}}} = {a^n}\) | Misol : \(\displaystyle \frac{1}{{{a^{ - 2}}}} = {a^2}\) |
| 9. \(\displaystyle \frac{{{a^{ - n}}}}{{{b^{ - m}}}} = \frac{{{b^m}}}{{{a^n}}}\) | Misol : \(\displaystyle \frac{{{a^{ - 6}}}}{{{b^{ - 17}}}} = \frac{{{b^{17}}}}{{{a^6}}}\) |
| 10. \({\left( {{a^n}{b^m}} \right)^k} = {a^{nk}}{b^{mk}}\) | Misol : \({\left( {{a^4}{b^{ - 9}}} \right)^3} = {a^{\left( 4 \right)\left( 3 \right)}}{b^{\left( { - 9} \right)\left( 3 \right)}} = {a^{12}}{b^{ - 27}}\) |
| 11. \(\displaystyle {\left( {\frac{{{a^n}}}{{{b^m}}}} \right)^k} = \frac{{{a^{nk}}}}{{{b^{mk}}}}\) | Misol : \({\left( {\displaystyle \frac{{{a^6}}}{{{b^5}}}} \right)^2} = \displaystyle \frac{{{a^{\left( 6 \right)\left( 2 \right)}}}}{{{b^{\left( 5 \right)\left( 2 \right)}}}} = \frac{{{a^{12}}}}{{{b^{10}}}}\) |
E'tibor bering, 4-xossa faqat ikkita ko'paytuvchi uchun yozilgan bo'lsa-da, uni istalgan sondagi ko'paytuvchilar uchun qo'llash mumkin, masalan: \({\left( {abcd} \right)^n} = {a^n}{b^n}{c^n}{d^n}\).
Eng ko'p uchraydigan odatiy xatolar
Talabalar daraja xossalarini qo'llashda ko'pincha quyidagi xatolarga yo'l qo'yishadi:
1-holat:
\[\begin{align*}& \text{To'g'ri : } & & a{b^{ - 2}} = a \cdot \frac{1}{{{b^2}}} = \frac{a}{{{b^2}}}\\ & \text{XATO : } & & a{b^{ - 2}} \ne \frac{1}{{a{b^2}}}\end{align*}\]Bu yerda manfiy daraja faqat \(b\) ga tegishli, chunki u darajaning bevosita chap tomonida turibdi. Shuning uchun faqat \(b\) maxrajga tushadi. \(a\) ko'paytuvchini maxrajga tushirish qat'iyan xatodir! Taqqoslash uchun: \({\left( {ab} \right)^{ - 2}} = \frac{1}{{{{\left( {ab} \right)}^2}}}\) bo'lgandagina qavs ichidagi ikkala harf ham maxrajga tushadi.
2-holat:
\[\begin{align*} & \text{To'g'ri : } & & \frac{1}{{3{a^{ - 5}}}} = \frac{1}{3} \cdot \frac{1}{{{a^{ - 5}}}} = \frac{{{a^5}}}{3}\\ & \text{XATO : } & & \frac{1}{{3{a^{ - 5}}}} \ne 3{a^5}\end{align*}\]Bu yerda \(-5\) daraja faqat \(a\) ga tegishli. 3 soni darajasiz maxrajda turibdi, shuning uchun u suratga chiqmaydi, maxrajda qoladi!
- \({\left( {4{x^{ - 4}}{y^5}} \right)^3}\)
- \({\left( { - 10{z^2}{y^{ - 4}}} \right)^2}{\left( {{z^3}y} \right)^{ - 5}}\)
- \(\displaystyle \frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}}\)
- \(\displaystyle \frac{{5{x^{ - 1}}{y^{ - 4}}}}{{{{\left( {3{y^5}} \right)}^{ - 2}}{x^9}}}\)
- \({\left( {\displaystyle \frac{{{z^{ - 5}}}}{{{z^{ - 2}}{x^{ - 1}}}}} \right)^6}\)
- \({\left( {\displaystyle \frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}}\)
a \({\left( {4{x^{ - 4}}{y^5}} \right)^3}\) Yechim
10-xossadan foydalanib, qavsdagi har bir hadni 3-darajaga ko'taramiz:
\[{\left( {4{x^{ - 4}}{y^5}} \right)^3} = {4^3}{x^{ - 12}}{y^{15}} = 64{x^{ - 12}}{y^{15}}\]Endi manfiy darajali \(x^{-12}\) ni maxrajga tushirib, musbat darajaga aylantiramiz:
\[= \frac{{64{y^{15}}}}{{{x^{12}}}}\]b \({\left( { - 10{z^2}{y^{ - 4}}} \right)^2}{\left( {{z^3}y} \right)^{ - 5}}\) Yechim
Ikkala qavsga ham 10-xossani qo'llaymiz:
\[{\left( { - 10} \right)^2}{z^4}{y^{ - 8}} \cdot {z^{ - 15}}{y^{ - 5}} = 100{z^{4 - 15}}{y^{ - 8 - 5}} = 100{z^{ - 11}}{y^{ - 13}}\]Manfiy ko'rsatkichli hadlarni maxrajga o'tkazamiz (100 soni suratda qoladi):
\[= \frac{{100}}{{{z^{11}}{y^{13}}}}\]c \(\displaystyle \frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}}\) Yechim
Manfiy darajali hadlarning o'rnini almashtiramiz: suratdagi \(n^{-2}\) maxrajga \(n^2\) bo'lib tushadi, maxrajdagi \(m^{-4}\) va \(n^{-3}\) esa suratga \(m^4\) va \(n^3\) bo'lib chiqadi:
\[\frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}} = \frac{{{m^4}{n^3}m}}{{7{n^2}}}\]Endi bir xil asosli darajalarni soddalashtiramiz:
\[= \frac{{{m^{4 + 1}}{n^{3 - 2}}}}{7} = \frac{{{m^5}n}}{7}\]d \(\displaystyle \frac{{5{x^{ - 1}}{y^{ - 4}}}}{{{{\left( {3{y^5}} \right)}^{ - 2}}{x^9}}}\) Yechim
Manfiy darajali ifodalarni ko'chiramiz. Maxrajdagi \((3y^5)^{-2}\) qavsli bo'lgani uchun, u butunligicha suratga musbat daraja bilan chiqadi:
\[\frac{{5{{\left( {3{y^5}} \right)}^2}}}{{x{y^4}{x^9}}} = \frac{{5 \cdot 9{y^{10}}}}{{{x^{1 + 9}}{y^4}}} = \frac{{45{y^{10 - 4}}}}{{{x^{10}}}} = \frac{{45{y^6}}}{{{x^{10}}}}\]e \({\left( {\displaystyle \frac{{{z^{ - 5}}}}{{{z^{ - 2}}{x^{ - 1}}}}} \right)^6}\) Yechim
Avval qavs ichidagi kasrni to'liq soddalashtirib olamiz:
\[{\left( {\frac{{{z^2}x^1}}{{{z^5}}}} \right)^6} = {\left( {\frac{x}{{{z^{5 - 2}}}}} \right)^6} = {\left( {\frac{x}{{{z^3}}}} \right)^6}\]Endi qavsdan tashqaridagi 6-darajani surat va maxrajga tatbiq etamiz:
\[= \frac{{{x^6}}}{{{z^{3 \cdot 6}}}} = \frac{{{x^6}}}{{{z^{18}}}}\]f \({\left( {\displaystyle \frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}}\) Yechim
Dastlab qavs ichini soddalashtiramiz (\(24/6 = 4\)):
\[{\left( {\frac{{4{a^3}{a^5}}}{{{b^8}b}}} \right)^{ - 2}} = {\left( {\frac{{4{a^8}}}{{{b^9}}}} \right)^{ - 2}}\]Tashqaridagi manfiy daraja tufayli kasrni to'ntaramiz (6-xossa) va darajani musbat qilamiz:
\[= {\left( {\frac{{{b^9}}}{{4{a^8}}}} \right)^2} = \frac{{{b^{18}}}}{{{4^2}{a^{16}}}} = \frac{{{b^{18}}}}{{16{a^{16}}}}\]Ushbu misollarda shart bo'yicha yakuniy javoblar faqat musbat darajalar bilan yozilishi talab qilindi. Bu yagona standart javob ko'rinishiga ega bo'lish uchun qilindi. Biroq keyingi darslarda va ayniqsa Calculus kursida manfiy darajalar bilan ishlash ko'pincha juda qulay va talab etiladigan shakl hisoblanadi.