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1.3-bo'lim: Ildizlar va radikallar (radicals)

Ushbu bo'limni radikal (ildiz) ta'rifidan boshlaymiz. Agar \(n\) — 1 dan katta musbat butun son va \(a\) — haqiqiy son bo'lsa, u holda:

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

Bu yerda \(n\) — ildiz ko'rsatkichi (index), \(a\) — ildizosti ifodasi (radicand), va \(\sqrt{\phantom{a}}\) belgisi esa radikal (ildiz belgisi) deb ataladi. Ushbu tenglikning chap tomoni radikal shakli, o'ng tomoni esa daraja shakli deb yuritiladi.

Ushbu ta'rifdan ko'rinib turibdiki, ildiz — bu avvalgi bo'limda o'rganganimiz ratsional (kasr) darajaning boshqacha yozilish shaklidir. Kvadrat ildiz uchun istisno tariqasida ildiz ko'rsatkichi odatda yozilmaydi:

\[\sqrt[2]{a} = \sqrt a \]
1-Misol Quyidagi ildizlarni kasr daraja ko'rinishida yozing:
  1. \(\sqrt[4]{{16}}\)
  2. \(\sqrt[{10}]{{8x}}\)
  3. \(\sqrt {{x^2} + {y^2}} \)
a \(\sqrt[4]{{16}}\) Yechim
\[\sqrt[4]{{16}} = {16^{\frac{1}{4}}}\]

b \(\sqrt[{10}]{{8x}}\) Yechim

Ildiz ostida butun boshli \(8x\) turgani sababli qavsdan foydalanish shart:

\[\sqrt[{10}]{{8x}} = {\left( {8x} \right)^{\frac{1}{{10}}}}\]

c \(\sqrt {{x^2} + {y^2}} \) Yechim
\[\sqrt {{x^2} + {y^2}} = {\left( {{x^2} + {y^2}} \right)^{\frac{1}{2}}}\]

MUHIM OGOHLANTIRISH: Hech qachon \(\sqrt{x^2+y^2}\) ni \(x + y\) ga teng deb hisoblamang! Ildiz yig'indi ustiga alohida-alohida taqsimlanmaydi!

2-Misol Quyidagi ildizlarning qiymatini hisoblang:
  1. \(\sqrt {16} \) va \(\sqrt[4]{{16}}\)
  2. \(\sqrt[5]{{243}}\)
  3. \(\sqrt[4]{{1296}}\)
  4. \(\sqrt[3]{{ - 125}}\)
  5. \(\sqrt[4]{{ - 16}}\)
a \(\sqrt {16} \) va \(\sqrt[4]{{16}}\) Yechim
\[\sqrt {16} = 4\hspace{0.25in}\text{va}\hspace{0.25in}\sqrt[4]{{16}} = 2\]

b \(\sqrt[5]{{243}}\) Yechim
\[\sqrt[5]{{243}} = 3\hspace{0.25in}\text{chunki}\hspace{0.25in}{3^5} = 243\]

c \(\sqrt[4]{{1296}}\) Yechim
\[\sqrt[4]{{1296}} = 6\hspace{0.25in}\text{chunki}\hspace{0.25in}{6^4} = 1296\]

d \(\sqrt[3]{{ - 125}}\) Yechim
\[\sqrt[3]{{ - 125}} = - 5\hspace{0.25in}\text{chunki}\hspace{0.25in}{\left( { - 5} \right)^3} = - 125\]

e \(\sqrt[4]{{ - 16}}\) Yechim

Haqiqiy sonlar sohasida yechim mavjud emas, chunki har qanday haqiqiy sonning to'rtinchi darajasi musbat bo'ladi, manfiy \(-16\) sonidan juft darajali haqiqiy ildiz chiqarib bo'lmaydi.

Ildizlarning asosiy xossalari (Properties of Radicals)

Agar \(n\) va \(m\) — musbat butun sonlar bo'lsa, quyidagi xossalar o'rinli:

  1. \(\sqrt[n]{{{a^n}}} = a\) (agar \(n\) toq son bo'lsa); va \(\sqrt[n]{{{a^n}}} = \left| a \right|\) (agar \(n\) juft son bo'lsa).
  2. \(\sqrt[n]{{ab}} = \sqrt[n]{a}\sqrt[n]{b}\)
  3. \(\displaystyle \sqrt[n]{{\frac{a}{b}}} = \frac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}}\)

1-xossadagi modul belgisiga alohida e'tibor bering: masalan, \(\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|\). Agar o'zgaruvchilar musbat deb berilgan bo'lsa, modul belgisini tashlab yuborish mumkin.

3-Misol Quyidagi ildizlarni soddalashtiring (barcha o'zgaruvchilarni musbat deb hisoblang):
  1. \(\sqrt {{y^7}} \)
  2. \(\sqrt[9]{{{x^6}}}\)
  3. \(\sqrt {18{x^6}{y^{11}}} \)
  4. \(\sqrt[4]{{32{x^9}{y^5}{z^{12}}}}\)
  5. \(\sqrt[5]{{\displaystyle \frac{{{x^6}}}{{{y^{11}}}}}}\)
a \(\sqrt {{y^7}} \) Yechim
\[\sqrt {{y^7}} = \sqrt {{y^6} \cdot y} = \sqrt {{y^6}} \sqrt y = {y^3}\sqrt y \]

b \(\sqrt[9]{{{x^6}}}\) Yechim
\[\sqrt[9]{{{x^6}}} = {x^{\frac{6}{9}}} = {x^{\frac{2}{3}}} = \sqrt[3]{{{x^2}}}\]

c \(\sqrt {18{x^6}{y^{11}}} \) Yechim
\[\sqrt {18{x^6}{y^{11}}} = \sqrt {9 \cdot 2 \cdot {x^6} \cdot {y^{10}} \cdot y} = 3{x^3}{y^5}\sqrt {2y} \]

d \(\sqrt[4]{{32{x^9}{y^5}{z^{12}}}}\) Yechim
\[\sqrt[4]{{16 \cdot 2 \cdot {x^8} \cdot x \cdot {y^4} \cdot y \cdot {z^{12}}}} = 2{x^2}y{z^3}\sqrt[4]{{2xy}}\]

e \(\sqrt[5]{{\displaystyle \frac{{{x^6}}}{{{y^{11}}}}}}\) Yechim
\[\sqrt[5]{{\frac{{{x^6}}}{{{y^{11}}}}}} = \frac{{\sqrt[5]{{{x^5} \cdot x}}}}{{\sqrt[5]{{{y^{10}} \cdot y}}}} = \frac{{x\sqrt[5]{x}}}{{{y^2}\sqrt[5]{y}}}\]
4-Misol Ko'paytirishni bajaring (\(x\) musbat deb hisoblansin):
  1. \(\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right)\)
  2. \(\left( {3\,\sqrt x - \sqrt y } \right)\left( {2\sqrt x - 5\sqrt y } \right)\)
  3. \(\left( {5\sqrt x + 2} \right)\left( {5\sqrt x - 2} \right)\)
a \(\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right)\) Yechim
\[\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right) = {\left( {\sqrt x } \right)^2} - 5\sqrt x + 2\sqrt x - 10 = x - 3\sqrt x - 10\]

b \(\left( {3\,\sqrt x - \sqrt y } \right)\left( {2\sqrt x - 5\sqrt y } \right)\) Yechim
\[= 6{\left( {\sqrt x } \right)^2} - 15\sqrt x \sqrt y - 2\sqrt x \sqrt y + 5{\left( {\sqrt y } \right)^2} = 6x - 17\sqrt {xy} + 5y\]

c \(\left( {5\sqrt x + 2} \right)\left( {5\sqrt x - 2} \right)\) Yechim

Kvadratlar ayirmasi formulasi \((a+b)(a-b) = a^2 - b^2\) dan foydalanamiz:

\[= {\left( {5\sqrt x } \right)^2} - {2^2} = 25x - 4\]

Maxrajni irratsionallikdan qutqarish (Rationalizing the Maxraj)

Kasr maxrajida ildiz qatnashsa, maxrajni butun songa yoki ratsional ifodaga keltirish uchun surat va maxraj mos qo'shma ko'paytuvchiga ko'paytiriladi.

5-Misol Quyidagi kasrlarning maxrajini irratsionallikdan qutqaring:
  1. \(\displaystyle \frac{4}{{\sqrt x }}\)
  2. \(\displaystyle \sqrt[5]{{\frac{2}{{{x^3}}}}}\)
  3. \(\displaystyle \frac{1}{{3 - \sqrt x }}\)
  4. \(\displaystyle \frac{4}{{2\sqrt x + \sqrt y }}\)
a \(\displaystyle \frac{4}{{\sqrt x }}\) Yechim

Surat va maxrajni \(\sqrt{x}\) ga ko'paytiramiz:

\[\frac{4}{{\sqrt x }} \cdot \frac{{\sqrt x }}{{\sqrt x }} = \frac{{4\sqrt x }}{x}\]

b \(\displaystyle \sqrt[5]{{\frac{2}{{{x^3}}}}}\) Yechim

Maxrajda \(x^3\) turibdi. 5-ildizdan to'liq chiqishi uchun \(x^5\) kerak, ya'ni yana \(x^2\) yetishmaydi. Surat va maxrajni \(\sqrt[5]{x^2}\) ga ko'paytiramiz:

\[\frac{{\sqrt[5]{2}}}{{\sqrt[5]{{{x^3}}}}} \cdot \frac{{\sqrt[5]{{{x^2}}}}}{{\sqrt[5]{{{x^2}}}}} = \frac{{\sqrt[5]{{2{x^2}}}}}{{\sqrt[5]{{{x^5}}}}} = \frac{{\sqrt[5]{{2{x^2}}}}}{x}\]

c \(\displaystyle \frac{1}{{3 - \sqrt x }}\) Yechim

Maxrajning qo'shmasi \(3 + \sqrt{x}\) ga ko'paytiramiz:

\[\frac{1}{{3 - \sqrt x }} \cdot \frac{{3 + \sqrt x }}{{3 + \sqrt x }} = \frac{{3 + \sqrt x }}{{{3^2} - {{\left( {\sqrt x } \right)}^2}}} = \frac{{3 + \sqrt x }}{{9 - x}}\]

d \(\displaystyle \frac{4}{{2\sqrt x + \sqrt y }}\) Yechim

Maxrajning qo'shmasi \(2\sqrt{x} - \sqrt{y}\) ga ko'paytiramiz:

\[\frac{4}{{2\sqrt x + \sqrt y }} \cdot \frac{{2\sqrt x - \sqrt y }}{{2\sqrt x - \sqrt y }} = \frac{{4\left( {2\sqrt x - \sqrt y } \right)}}{{{{\left( {2\sqrt x } \right)}^2} - {{\left( {\sqrt y } \right)}^2}}} = \frac{{8\sqrt x - 4\sqrt y }}{{4x - y}}\]