n-darajali ildiz va ratsional ko'rsatkichli daraja

Bu nima va nega kerak

Kvadratning yuzidan tomonini topish uchun kvadrat ildiz ($\sqrt{S}$), kubning hajmidan qirrasini topish uchun esa kub ildiz ($\sqrt[3]{V}$) ishlatiladi.

Fizika, biologiya va moliyaviy hisob-kitoblarda jarayonlar faqat butun darajalar orqali emas, balki kasr ko'rsatkichli darajalar orqali ifodalanadi. Masalan, tana massasi va metabolizm tezligi o'rtasidagi Kleyber qonuni $B \propto M^{\frac{3}{4}}$ ko'rinishida yoziladi.

Ildiz va daraja amallarini yagona qoidaga birlashtirish uchun ratsional ko'rsatkichli daraja va n-darajali ildiz tushunchalari kiritiladi.

Asosiy g'oya

1. n-darajali ildiz ta'rifi

$n \in \mathbb{N}, n \ge 2$ bo'lsin. $n$-darajasi $a$ ga teng bo'lgan son $a$ ning $n$-darajali ildizi deyiladi ($x^n = a$).

• Juft darajali ildiz ($n = 2k$):

  • Manfiy sondan juft darajali ildiz mavjud emas.
  • $a \ge 0$ bo'lganda, $n$-darajasi $a$ ga teng bo'lgan nomanfiy son arifmetik ildiz deyiladi va $\sqrt[2k]{a}$ deb yoziladi ($\sqrt[2k]{a} \ge 0$).
  • Eng muhim ayniyat: $\sqrt[2k]{a^{2k}} = |a|$.

• Toq darajali ildiz ($n = 2k + 1$):

  • Ixtiyoriy haqiqiy son $a \in \mathbb{R}$ dan mavjud.
  • $\sqrt[2k+1]{-a} = -\sqrt[2k+1]{a}$.
  • Toq darajali ildiz uchun: $\sqrt[2k+1]{a^{2k+1}} = a$.

2. n-darajali ildizning asosiy xossalari

$a \ge 0, b > 0, k, m, n \in \mathbb{N}$ bo'lganda:

1. $\sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b}$
2. $\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$
3. $(\sqrt[n]{a})^k = \sqrt[n]{a^k}$
4. $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$
5. $\sqrt[kn]{a^{km}} = \sqrt[n]{a^m}$

3. Ratsional ko'rsatkichli daraja

Ta'rif. $a > 0$ va $r = \dfrac{m}{n}$ ($m \in \mathbb{Z}, n \in \mathbb{N}, n \ge 2$) ratsional son bo'lsa: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ $a^{-\frac{m}{n}} = \frac{1}{a^{\frac{m}{n}}} = \frac{1}{\sqrt[n]{a^m}}$

Nega $a > 0$ sharti qo'yiladi? Chunki manfiy asosda kasrlarni turlicha yozish ziddiyatga olib keladi: masalan, $(-8)^{\frac{1}{3}} = \sqrt[3]{-8} = -2$, lekin $\dfrac{1}{3} = \dfrac{2}{6}$ bo'lgani uchun $(-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2$. Bunday qarama-qarshilikning oldini olish uchun ratsional ko'rsatkichli darajaning asosi doimo $a > 0$ deb olinadi.

4. Ratsional darajaning xossalari

$a > 0, b > 0$ va ixtiyoriy $p, q \in \mathbb{Q}$ uchun:

1. $a^p \cdot a^q = a^{p+q}$
2. $a^p : a^q = a^{p-q}$
3. $(a^p)^q = a^{pq}$
4. $(a \cdot b)^p = a^p \cdot b^p$
5. $\left(\dfrac{a}{b}\right)^p = \dfrac{a^p}{b^p}$

Yechilgan misollar

1-misol. Sonli ildizli ifodalarni hisoblash

Hisoblang: $A = \sqrt[3]{54} \cdot \sqrt[3]{4} - \frac{\sqrt[4]{162}}{\sqrt[4]{2}} + \sqrt[5]{-32}$

Yechilishi:

1. Ildizlarni ko'paytirish xossasi: $\sqrt[3]{54} \cdot \sqrt[3]{4} = \sqrt[3]{54 \cdot 4} = \sqrt[3]{216} = 6$
2. Ildizlarni bo'lish xossasi: $\frac{\sqrt[4]{162}}{\sqrt[4]{2}} = \sqrt[4]{\frac{162}{2}} = \sqrt[4]{81} = 3$
3. Toq darajali manfiy ildiz: $\sqrt[5]{-32} = -2$
4. Umumiy qiymat: $A = 6 - 3 + (-2) = 1$

Javob: $1$.

2-misol. O'zgaruvchili ifodalarni soddalashtirish

Agar $x \le 2$ bo'lsa, ifodani soddalashtiring: $B = \sqrt[4]{(x - 2)^4} + \sqrt[3]{(x - 5)^3}$

Yechilishi:

1. Juft ko'rsatkichli ildizdan modul chiqadi: $\sqrt[4]{(x - 2)^4} = |x - 2|$ Berilgan shartga ko'ra $x \le 2$, ya'ni $x - 2 \le 0$. Demak: $|x - 2| = -(x - 2) = 2 - x$
2. Toq ko'rsatkichli ildizdan ifodaning o'zi chiqadi: $\sqrt[3]{(x - 5)^3} = x - 5$
3. Yig'indini topamiz: $B = (2 - x) + (x - 5) = -3$

Javob: $-3$.

3-misol. Ratsional ko'rsatkichli darajalar ustida amallar

Kasrni qisqartiring: $C = \frac{a - b}{a^{\frac{1}{2}} + b^{\frac{1}{2}}} - \frac{a^{\frac{3}{2}} - b^{\frac{3}{2}}}{a - b}$ ($a > 0, b > 0, a \neq b$).

Yechilishi:

1. Birinchi kasr suratini kvadratlar ayirmasi kabi yoyamiz: $a - b = (a^{\frac{1}{2}})^2 - (b^{\frac{1}{2}})^2 = (a^{\frac{1}{2}} - b^{\frac{1}{2}})(a^{\frac{1}{2}} + b^{\frac{1}{2}})$ Kasr qisqaradi: $a^{\frac{1}{2}} - b^{\frac{1}{2}}$.

2. Ikkinchi kasr suratini kublar ayirmasi kabi yoyamiz: $a^{\frac{3}{2}} - b^{\frac{3}{2}} = (a^{\frac{1}{2}} - b^{\frac{1}{2}})(a + a^{\frac{1}{2}}b^{\frac{1}{2}} + b)$ Maxraj: $a - b = (a^{\frac{1}{2}} - b^{\frac{1}{2}})(a^{\frac{1}{2}} + b^{\frac{1}{2}})$. Qisqargandan so'ng umumiy maxrajga keltirish o'rniga to'g'ridan-to'g'ri hisoblash ham mumkin: $C = (a^{\frac{1}{2}} - b^{\frac{1}{2}}) - \frac{a + a^{\frac{1}{2}}b^{\frac{1}{2}} + b}{a^{\frac{1}{2}} + b^{\frac{1}{2}}}$ Umumiy maxraj $a^{\frac{1}{2}} + b^{\frac{1}{2}}$: $\text{Surat} = (a^{\frac{1}{2}} - b^{\frac{1}{2}})(a^{\frac{1}{2}} + b^{\frac{1}{2}}) - (a + a^{\frac{1}{2}}b^{\frac{1}{2}} + b) = a - b - a - a^{\frac{1}{2}}b^{\frac{1}{2}} - b = -2b - a^{\frac{1}{2}}b^{\frac{1}{2}}$

Javob: $-\dfrac{2b + \sqrt{ab}}{\sqrt{a} + \sqrt{b}}$.

4-misol. Maxrajdagi irratsionallikni yo'qotish

Kasr maxrajidagi irratsionallikni yo'qoting: a) $\dfrac{6}{\sqrt[3]{4}}$ b) $\dfrac{1}{\sqrt[3]{3} - 1}$

Yechilishi: a) $\sqrt[3]{4} = \sqrt[3]{2^2}$. To'la kub qilish uchun surat va maxrajni $\sqrt[3]{2}$ ga ko'paytiramiz: $\frac{6}{\sqrt[3]{4}} = \frac{6 \cdot \sqrt[3]{2}}{\sqrt[3]{2^2} \cdot \sqrt[3]{2}} = \frac{6\sqrt[3]{2}}{\sqrt[3]{8}} = \frac{6\sqrt[3]{2}}{2} = 3\sqrt[3]{2}.$

b) $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ formulasidan foydalanamiz. Surat va maxrajni $\sqrt[3]{3^2} + \sqrt[3]{3} \cdot 1 + 1^2 = \sqrt[3]{9} + \sqrt[3]{3} + 1$ ga ko'paytiramiz: $\frac{1}{\sqrt[3]{3} - 1} = \frac{\sqrt[3]{9} + \sqrt[3]{3} + 1}{(\sqrt[3]{3})^3 - 1^3} = \frac{\sqrt[3]{9} + \sqrt[3]{3} + 1}{3 - 1} = \frac{\sqrt[3]{9} + \sqrt[3]{3} + 1}{2}.$

Javob: a) $3\sqrt[3]{2}$; b) $\dfrac{\sqrt[3]{9} + \sqrt[3]{3} + 1}{2}$.

Mashq

1. Hisoblang: $\sqrt[3]{-125} + \sqrt[4]{81} - \sqrt[5]{-32}$.
2. Hisoblang: $\sqrt[3]{16} \cdot \sqrt[3]{4}$.
3. Hisoblang: $\dfrac{\sqrt[4]{48}}{\sqrt[4]{3}}$.
4. Hisoblang: $27^{\frac{2}{3}} + 16^{-\frac{3}{4}}$.
5. Soddalashtiring: $\sqrt[6]{a^2 \sqrt{a}}$ ($a > 0$).
6. $x < 1$ bo'lganda $\sqrt[4]{(x - 1)^4} + \sqrt{(x - 4)^2}$ ifodaning qiymatini toping.
7. Kasrni qisqartiring: $\dfrac{a - 9}{a^{\frac{1}{2}} + 3}$.
8. Hisoblang: $(2^{\sqrt{3}})^{\sqrt{3}} \cdot 4^{-0{,}5}$.
9. Maxrajdagi irratsionallikni yo'qoting: $\dfrac{4}{\sqrt[3]{2}}$.
10. Soddalashtiring: $\dfrac{x^{\frac{3}{2}} - y^{\frac{3}{2}}}{x^{\frac{1}{2}} - y^{\frac{1}{2}}} - x^{\frac{1}{2}} y^{\frac{1}{2}}$ ($x > 0, y > 0, x \neq y$).

Javoblar va yechim yo'llari

1. $-5 + 3 - (-2) = -5 + 3 + 2 = 0$. Javob: $0$.

2. $\sqrt[3]{16 \cdot 4} = \sqrt[3]{64} = 4$. Javob: $4$.

3. $\sqrt[4]{\dfrac{48}{3}} = \sqrt[4]{16} = 2$. Javob: $2$.

4. $27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 9$; $16^{-\frac{3}{4}} = (2^4)^{-\frac{3}{4}} = 2^{-3} = \dfrac{1}{8}$. $9 + \dfrac{1}{8} = \dfrac{73}{8} = 9{,}125$. Javob: $9{,}125$ (yoki $\dfrac{73}{8}$).

5. $\sqrt{a} = a^{\frac{1}{2}}$. $a^2 \cdot a^{\frac{1}{2}} = a^{\frac{5}{2}}$. $\sqrt[6]{a^{\frac{5}{2}}} = (a^{\frac{5}{2}})^{\frac{1}{6}} = a^{\frac{5}{12}}$. Javob: $a^{\frac{5}{12}}$.

6. $x < 1$ bo'lsa, $|x - 1| = 1 - x$ va $|x - 4| = 4 - x$. Natija: $(1 - x) + (4 - x) = 5 - 2x$. Javob: $5 - 2x$.

7. $a - 9 = (a^{\frac{1}{2}} - 3)(a^{\frac{1}{2}} + 3)$. Kasr $a^{\frac{1}{2}} - 3$ ga teng. Javob: $a^{\frac{1}{2}} - 3$ (yoki $\sqrt{a} - 3$).

8. $(2^{\sqrt{3}})^{\sqrt{3}} = 2^3 = 8$. $4^{-0{,}5} = (2^2)^{-0{,}5} = 2^{-1} = 0{,}5$. $8 \cdot 0{,}5 = 4$. Javob: $4$.

9. $\dfrac{4\sqrt[3]{4}}{\sqrt[3]{2} \cdot \sqrt[3]{4}} = \dfrac{4\sqrt[3]{4}}{2} = 2\sqrt[3]{4}$. Javob: $2\sqrt[3]{4}$.

10. Surat: $(x^{\frac{1}{2}} - y^{\frac{1}{2}})(x + x^{\frac{1}{2}}y^{\frac{1}{2}} + y)$. Bo'lganda: $x + x^{\frac{1}{2}}y^{\frac{1}{2}} + y$. Ayirganda: $(x + x^{\frac{1}{2}}y^{\frac{1}{2}} + y) - x^{\frac{1}{2}}y^{\frac{1}{2}} = x + y$. Javob: $x + y$.

Keng tarqalgan xatolar

1. Juft darajali ildizdan modul chiqishini unutish. ✗ $\sqrt[4]{a^4} = a$ deb yozish. ✓ Agar $a < 0$ bo'lsa, bu noto'g'ri. Doimo $\sqrt[4]{a^4} = |a|$ bo'ladi. Masalan, $\sqrt[4]{(-3)^4} = |-3| = 3$.

2. Manfiy asosni ratsional darajaga oshirish. ✗ $(-4)^{\frac{1}{2}} = \sqrt{-4} = -2$. ✓ Ratsional ko'rsatkichli daraja faqat musbat asoslar ($a > 0$) uchun aniqlangan. $(-4)^{\frac{1}{2}}$ ma'noga ega emas.

3. Ildiz ostidagi yig'indini bo'lib yuborish. ✗ $\sqrt[3]{a^3 + b^3} = a + b$. ✓ Ildiz faqat ko'paytma va bo'linmaga nisbatan tarqaladi. Yig'indidan alohida ildiz chiqarib bo'lmaydi.

4. Kasr ko'rsatkichining surat va maxrajini adashtirish. ✗ $a^{\frac{3}{2}} = \sqrt[3]{a^2}$ deb yozish. ✓ Maxraj ildiz ko'rsatkichidir, surat esa darajadir: $a^{\frac{3}{2}} = \sqrt{a^3}$.

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Sonlar va darajalar nazariyasini yakunladik. Keyingi katta bo'lim — Trigonometriya bo'lib, uning birinchi darsida burchakning yangi va universal o'lchov birligi — radian o'lchovi bilan tanishamiz.