n-darajali ildiz va ratsional ko'rsatkichli daraja: matematika darsi

Hajmi $27$ kub santimetr bo'lgan kubning qirrasi $\sqrt[3]{27}=3$ santimetr. Xuddi shunday, $8^{\frac{2}{3}}$ yozuvi ham ildizning boshqacha ko'rinishi: u $\sqrt[3]{8^2}=4$ ga teng. Bu darsda istalgan darajali ildiz, uning xossalari va kasr ko'rsatkichli darajalar bilan ishlash o'rganiladi.

n-darajali ildiz

$n\ge 2$ natural son bo'lsin. $a$ sonning $n$ -darajali ildizi deb $n$ -darajasi $a$ ga teng bo'lgan songa aytiladi. Juft $n$ da faqat manfiy bo'lmagan qiymat olinadi.

Xossa	Toq $n$	Juft $n$
XossaQaysi $a$ uchun mavjud	Toq $n$ har qanday $a$	Juft $n$ faqat $a\ge 0$
XossaQiymat ishorasi	Toq $n$ $a$ ning ishorasi bilan bir xil	Juft $n$ manfiy emas
XossaMisol	Toq $n$ $\sqrt[3]{-27}=-3$ , $\sqrt[5]{-32}=-2$	Juft $n$ $\sqrt[4]{16}=2$ , $\sqrt[4]{-16}$ mavjud emas
Xossa $x^n=a$ yechimlari	Toq $n$ bitta: $x=\sqrt[n]{a}$	Juft $n$ $a>0$ da ikkita: $x=\pm\sqrt[n]{a}$ ; $a=0$ da bitta: $x=0$ ; $a<0$ da yo'q

Quyidagi darajalar ko'p kerak bo'ladi. Katta sonlarni ham shu tarzda taniymiz: $2187=3^7$ , $1024=2^{10}$ .

Asos	$n=2$	$n=3$	$n=4$
Asos $2$	$n=2$ $4$	$n=3$ $8$	$n=4$ $16$
Asos $3$	$n=2$ $9$	$n=3$ $27$	$n=4$ $81$
Asos $4$	$n=2$ $16$	$n=3$ $64$	$n=4$ $256$
Asos $5$	$n=2$ $25$	$n=3$ $125$	$n=4$ $625$
Asos $7$	$n=2$ $49$	$n=3$ $343$	$n=4$ $2401$

Asos	$n=5$	$n=6$
Asos $2$	$n=5$ $32$	$n=6$ $64$
Asos $3$	$n=5$ $243$	$n=6$ $729$
Asos $4$	$n=5$ $1024$	$n=6$ $4096$
Asos $5$	$n=5$ $3125$	$n=6$ $15625$
Asos $7$	$n=5$ $16807$	$n=6$ $117649$

Darajaning ildizi: toq va juft

\sqrt[n]{a^n}=a\ \ (n\ \text{toq})

\sqrt[n]{a^n}=|a|\ \ (n\ \text{juft})

Juft darajali ildiz manfiy bo'lmaydi, shuning uchun modul paydo bo'ladi

Ifoda	Hisob	Qiymat
Ifoda $\sqrt[3]{(-4)^3}$	Hisobtoq daraja, son o'zgarmaydi	Qiymat $-4$
Ifoda $\sqrt[3]{-4^3}$	Hisob $-4^3=-64$ , $\sqrt[3]{-64}$	Qiymat $-4$
Ifoda $\sqrt[4]{(-3)^4}$	Hisob $\|-3\|$	Qiymat $3$
Ifoda $\sqrt[6]{(1-\sqrt{2})^6}$	Hisob $\|1-\sqrt{2}\|$ , $\sqrt{2}>1$	Qiymat $\sqrt{2}-1$

Ildiz xossalari

Juft $n$ da quyidagilar $a\ge 0$ , $b\ge 0$ uchun o'rinli, toq $n$ da birinchi ikkisi har qanday ishorada ishlaydi. Maxraj nolga teng bo'lmasligi shart.

Xossa	Misol
$\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}$	$\sqrt[3]{4}\cdot\sqrt[3]{16}=\sqrt[3]{64}=4$
$\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$	$\dfrac{\sqrt[3]{54}}{\sqrt[3]{2}}=\sqrt[3]{27}=3$
$(\sqrt[n]{a})^k=\sqrt[n]{a^k}$	$(\sqrt[3]{8})^2=\sqrt[3]{64}=4$
$\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$	$\sqrt[3]{\sqrt{64}}=\sqrt[6]{64}=2$
$\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}$	$\sqrt[4]{25}=\sqrt{5}$ , $\sqrt[6]{8}=\sqrt{2}$

Ichma-ich ildizlarda darajalar ko'paytiriladi va son toza chiqadi: $\sqrt[3]{\sqrt{\dfrac{1}{729}}}=\sqrt[6]{\dfrac{1}{3^6}}=\dfrac{1}{3}$ .

Daraja ko'rsatkichi bilan ildiz darajasini qisqartirish faqat $a\ge 0$ da to'g'ri: $\sqrt[6]{(-2)^2}=\sqrt[6]{4}=\sqrt[3]{2}$ , $\sqrt[3]{-2}$ emas.

Ildiz ostidan chiqarish va o'xshash ildizlar

\sqrt[n]{a^nb}=a\sqrt[n]{b}\ \ (a,b\ge 0)

\sqrt[4]{a^4b}=|a|\sqrt[4]{b}\ \ (b\ge 0)

Kub ildizda to'liq kub, to'rtinchi darajali ildizda to'rtinchi daraja ajratiladi. Toq

n

a

b

istalgan ishorali bo'lishi mumkin

\sqrt[3]{24}+\sqrt[3]{81}-\sqrt[3]{192}

=2\sqrt[3]{3}+3\sqrt[3]{3}-4\sqrt[3]{3}=\sqrt[3]{3}

O'xshash ildizlar: darajasi ham, ildiz ostisi ham bir xil

Kasr ildiz ostida bo'lsa, maxrajni to'liq kubga to'ldiramiz. $\sqrt[3]{\dfrac{1}{2}}=\sqrt[3]{\dfrac{4}{8}}=\dfrac{\sqrt[3]{4}}{2}$ , $\sqrt[3]{\dfrac{4}{27}}=\dfrac{\sqrt[3]{4}}{3}$ , $\sqrt[3]{32}=2\sqrt[3]{4}$ . Shuning uchun:

4\sqrt[3]{\frac{1}{2}}+6\sqrt[3]{\frac{4}{27}}-\sqrt[3]{32}

=2\sqrt[3]{4}+2\sqrt[3]{4}-2\sqrt[3]{4}=2\sqrt[3]{4}

Har xil darajali ildizlar

Ildiz darajalarini umumiy songa, ya'ni ularning EKUKiga keltiramiz. Bu ildizlarni ko'paytirish va solishtirish uchun kerak:

\sqrt{2}\cdot\sqrt[3]{3}=\sqrt[6]{2^3}\cdot\sqrt[6]{3^2}=\sqrt[6]{72}

2

3

ning EKUKi

6

Solishtirish: $\sqrt{2}=\sqrt[6]{8}$ , $\sqrt[3]{3}=\sqrt[6]{9}$ , demak $\sqrt{2}<\sqrt[3]{3}$ .

Musbat son uchun oltinchi darajali ildiz kub ildizning kvadrat ildizi. $3-\sqrt{2}$ musbat bo'lgani uchun $\sqrt[6]{(3-\sqrt{2})^2}=\sqrt[3]{3-\sqrt{2}}$ . To'rtinchi darajali ildiz ikki marta kvadrat ildiz, shuning uchun u ikki marta ochiladi:

\sqrt[4]{97+56\sqrt{3}}

=\sqrt{\sqrt{97+2\sqrt{2352}}}

=\sqrt{7+4\sqrt{3}}=2+\sqrt{3}

49+48=97

49\cdot 48=2352

; keyin

7+4\sqrt{3}=(2+\sqrt{3})^2

Kub ildizni oltinchi darajali ildizga keltirish uchun ildiz ostini kvadratga oshiramiz: $a,b\ge 0$ da $\sqrt[3]{a}\cdot\sqrt[6]{b}=\sqrt[6]{a^2b}$ . Masalan, $(1+\sqrt{3})^2=4+2\sqrt{3}$ , shuning uchun:

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