Ratsional ko'rsatkichli daraja va uning xossalari

Ushbu darsda daraja tushunchasini yanada kengaytiramiz: ko‘rsatkich endi butun son emas, balki kasr (ratsional son) bo‘lishi mumkin. Bu bizni ildizlar dunyosi bilan bog‘laydi va $\sqrt[n]{a}$ kabi ifodalarni daraja tilida yozish imkonini beradi.

§1. Ratsional ko‘rsatkichli daraja tushunchasi

$a^2=9$ tenglamaning yechimi $a=\sqrt{9}=3$ edi. Ildiz — darajaga teskari amal. Endi savol: $a^{1/2}$ nimani anglatadi? Uni shunday aniqlaymizki, darajaning eski xossalari saqlansin.

Ta’rif. $a>0$ va $n$ natural son bo‘lsa:

$$a^{\frac{1}{n}} = \sqrt[n]{a}$$

Umuman, $m$ butun, $n$ natural ($n>1$) bo‘lganda:

$$a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}$$

Nega aynan shunday? Agar $a^{1/2}$ ni $\sqrt{a}$ deb olsak, u holda $\left(a^{1/2}\right)^2 = a^{\frac{1}{2}\cdot 2}=a^1=a$ bo‘ladi — bu esa ildizning ta’rifiga to‘liq mos.

Misol. $27^{\frac{1}{3}}$ va $16^{\frac{3}{4}}$ ni hisoblang.
Yechish.

$$27^{\frac{1}{3}} = \sqrt[3]{27} = 3$$

$$16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8$$

Javob. $3$ va $8$.

Eslatma. Ratsional ko‘rsatkichli daraja odatda musbat asos uchun aniqlanadi ($a>0$). Sababi: $(-8)^{1/3}$ kabi ifodalarda kasrni qisqartirsak ziddiyat kelib chiqadi. Shu bois asos manfiy bo‘lganda ehtiyot bo‘lish kerak.

§2. Xossalarning saqlanishi

Eng muhim jihat shuki, butun ko‘rsatkichlar uchun o‘rgangan barcha xossalar ratsional ko‘rsatkichlar uchun ham to‘liq kuchda qoladi.

Qoida. $a>0,\ b>0$ va $p,q$ ixtiyoriy ratsional sonlar uchun:

$$a^{p}\cdot a^{q}=a^{p+q}, \qquad \dfrac{a^{p}}{a^{q}}=a^{p-q}, \qquad (a^{p})^{q}=a^{pq}$$

$$(ab)^{p}=a^{p}b^{p}, \qquad \left(\dfrac{a}{b}\right)^{p}=\dfrac{a^{p}}{b^{p}}$$

Misol. $a^{\frac{1}{2}}\cdot a^{\frac{1}{3}}$ ni soddalashtiring.
Yechish. Ko‘rsatkichlarni umumiy maxrajga keltirib qo‘shamiz:

$$a^{\frac{1}{2}}\cdot a^{\frac{1}{3}} = a^{\frac{1}{2}+\frac{1}{3}} = a^{\frac{3+2}{6}} = a^{\frac{5}{6}}$$

Javob. $a^{\frac{5}{6}}$.

Misol. $\dfrac{x^{\frac{3}{4}}}{x^{\frac{1}{4}}}$ ni soddalashtiring.
Yechish.

$$\dfrac{x^{\frac{3}{4}}}{x^{\frac{1}{4}}} = x^{\frac{3}{4}-\frac{1}{4}} = x^{\frac{2}{4}} = x^{\frac{1}{2}} = \sqrt{x}$$

Javob. $\sqrt{x}$.

§3. Ildizlarni daraja orqali soddalashtirish

Ratsional ko‘rsatkichning asosiy foydasi shundaki, chalkash ildizli ifodalarni daraja tiliga o‘tkazib, oson ishlash mumkin.

Savol. $\sqrt{a}\cdot\sqrt[3]{a}$ ni bitta ildiz ostida yozish qiyinmi? Daraja tilida esa bu bir necha soniyalik ish.

Misol. $\sqrt{a}\cdot\sqrt[3]{a}$ ni soddalashtiring.
Yechish. Har bir ildizni darajaga aylantiramiz:

$$\sqrt{a}\cdot\sqrt[3]{a} = a^{\frac{1}{2}}\cdot a^{\frac{1}{3}} = a^{\frac{5}{6}} = \sqrt[6]{a^{5}}$$

Javob. $\sqrt[6]{a^{5}}$.

Misol. $\sqrt[3]{\,8^{2}\,}$ ni hisoblang.
Yechish.

$$\sqrt[3]{8^{2}} = 8^{\frac{2}{3}} = \left(8^{\frac{1}{3}}\right)^{2} = 2^{2} = 4$$

Javob. $4$.

§ Lug‘at

ratsional ko‘rsatkichli daraja — rational exponent power
ildiz — root / radical
$n$-darajali ildiz — $n$-th root
kvadrat ildiz — square root
kub ildiz — cube root
ildiz ostidagi ifoda — radicand
kasr ko‘rsatkich — fractional exponent