✏️ Differensial Hisob — Masalalar

1-bo'lim: Limitlar (5 ta masala)

Masala 1.1 ⭐⭐

Limitni hisoblang: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Yechim

Qadam 1: To'g'ridan-to'g'ri qo'ysak $\frac{0}{0}$ noaniqlik.

Qadam 2: Faktorlaymiz: $\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$

Qadam 3: Limit: $\lim_{x \to 2} (x + 2) = 4$

Javob: $4$

Masala 1.2 ⭐⭐

$\lim_{x \to 0} \frac{\sin 3x}{x}$

Yechim

Usul: $\frac{\sin u}{u} \to 1$ formulasini ishlatamiz.

$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$

Javob: $3$

Masala 1.3 ⭐⭐⭐

L'Hôpital qoidasi bilan: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Yechim

Tekshirish: $x = 0$ da $\frac{0}{0}$ noaniqlik.

1-L'Hôpital: $\lim_{x \to 0} \frac{e^x - 1}{2x} = \frac{0}{0}$

Hali noaniqlik, yana qo'llaymiz:

2-L'Hôpital: $\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$

Javob: $\frac{1}{2}$

Masala 1.4 ⭐⭐⭐

$\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^x$

Yechim

Usul: $(1 + \frac{1}{n})^n \to e$ formulasini moslashtiramiz.

$u = \frac{x}{3}$ almashtiruv: $\left(1 + \frac{3}{x}\right)^x = \left(1 + \frac{1}{u}\right)^{3u} = \left[\left(1 + \frac{1}{u}\right)^u\right]^3$

$x \to \infty$ da $u \to \infty$: $\lim_{u \to \infty} \left[\left(1 + \frac{1}{u}\right)^u\right]^3 = e^3$

Javob: $e^3 \approx 20.09$

Masala 1.5 ⭐⭐⭐⭐

$\lim_{x \to 0} \frac{\tan x - x}{x - \sin x}$

Yechim

L'Hôpital (3 marta):

$\lim \frac{\sec^2 x - 1}{1 - \cos x} = \lim \frac{\tan^2 x}{1 - \cos x}$

Taylor yoyilmasi:

• $\tan x \approx x + \frac{x^3}{3}$
• $\tan^2 x \approx x^2$
• $1 - \cos x \approx \frac{x^2}{2}$

$\lim_{x \to 0} \frac{x^2}{x^2/2} = 2$

Javob: $2$

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2-bo'lim: Oddiy hosilalar (6 ta masala)

Masala 2.1 ⭐⭐

Hosilani toping: $f(x) = 3x^4 - 2x^3 + 5x - 7$

Yechim

$f'(x) = 12x^3 - 6x^2 + 5$

Javob: $f'(x) = 12x^3 - 6x^2 + 5$

Masala 2.2 ⭐⭐

$f(x) = e^{2x} \cdot \sin x$

Yechim

Ko'paytma qoidasi: $(uv)' = u'v + uv'$

• $u = e^{2x}$, $u' = 2e^{2x}$
• $v = \sin x$, $v' = \cos x$

$f'(x) = 2e^{2x}\sin x + e^{2x}\cos x = e^{2x}(2\sin x + \cos x)$

Javob: $f'(x) = e^{2x}(2\sin x + \cos x)$

Masala 2.3 ⭐⭐

$f(x) = \frac{x^2 + 1}{x^2 - 1}$

Yechim

Bo'linma qoidasi: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

• $u = x^2 + 1$, $u' = 2x$
• $v = x^2 - 1$, $v' = 2x$

$f'(x) = \frac{2x(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2}$

$= \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}$

Javob: $f'(x) = \frac{-4x}{(x^2-1)^2}$

Masala 2.4 ⭐⭐⭐

$f(x) = \ln(\cos x)$

Yechim

Zanjir qoidasi: $(f(g))' = f'(g) \cdot g'$

$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$

Javob: $f'(x) = -\tan x$

Masala 2.5 ⭐⭐⭐

$f(x) = \arctan(e^x)$

Yechim

Zanjir qoidasi:

$f'(x) = \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{e^x}{1 + e^{2x}}$

Javob: $f'(x) = \frac{e^x}{1 + e^{2x}}$

Masala 2.6 ⭐⭐⭐

$f(x) = x^x$ ($x \gt 0$)

Yechim

Logarifmik differensiallash:

$\ln f = x \ln x$

Ikkala tomonni differensiallaymiz: $\frac{f'}{f} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

$f' = f(\ln x + 1) = x^x(\ln x + 1)$

Javob: $f'(x) = x^x(\ln x + 1)$

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3-bo'lim: Yuqori tartibli hosilalar (4 ta masala)

Masala 3.1 ⭐⭐

$f(x) = x^5 - 3x^3 + 2x$ uchun $f''(x)$ va $f'''(x)$ ni toping.

Yechim

$f'(x) = 5x^4 - 9x^2 + 2$ $f''(x) = 20x^3 - 18x$ $f'''(x) = 60x^2 - 18$

Javob: $f''(x) = 20x^3 - 18x$, $f'''(x) = 60x^2 - 18$

Masala 3.2 ⭐⭐⭐

$f(x) = e^{ax}$ uchun $f^{(n)}(x)$ ni toping.

Yechim

$f'(x) = ae^{ax}$ $f''(x) = a^2e^{ax}$ $f'''(x) = a^3e^{ax}$

Qonuniyat: $f^{(n)}(x) = a^n e^{ax}$

Javob: $f^{(n)}(x) = a^n e^{ax}$

Masala 3.3 ⭐⭐⭐

$f(x) = \sin x$ uchun $f^{(100)}(x)$ ni toping.

Yechim

$f'(x) = \cos x$ $f''(x) = -\sin x$ $f'''(x) = -\cos x$ $f^{(4)}(x) = \sin x$

Davr = 4. $100 = 4 \cdot 25 + 0$

$f^{(100)}(x) = f^{(0)}(x) = \sin x$

Javob: $f^{(100)}(x) = \sin x$

Masala 3.4 ⭐⭐⭐⭐

Robot qo'li harakati: $s(t) = 2t^3 - 9t^2 + 12t + 5$ (m).

a) Tezlik va tezlanishni toping b) Robot qachon to'xtaydi? c) Jerk qancha?

Yechim

a) Tezlik va tezlanish: $v(t) = s'(t) = 6t^2 - 18t + 12$ $a(t) = v'(t) = 12t - 18$

b) To'xtash ($v = 0$): $6t^2 - 18t + 12 = 0$ $t^2 - 3t + 2 = 0$ $(t-1)(t-2) = 0$ $t = 1s \text{ va } t = 2s$

c) Jerk: $j(t) = a'(t) = 12 \text{ m/s}^3$

Javob:

• a) $v = 6t^2 - 18t + 12$, $a = 12t - 18$
• b) $t = 1s$ va $t = 2s$
• c) $j = 12$ m/s³ (konstant)

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4-bo'lim: Qisman hosilalar (5 ta masala)

Masala 4.1 ⭐⭐

$f(x, y) = x^3y + 2xy^2 - 3x + y$ uchun $\frac{\partial f}{\partial x}$ va $\frac{\partial f}{\partial y}$ ni toping.

Yechim

$\frac{\partial f}{\partial x} = 3x^2y + 2y^2 - 3$

$\frac{\partial f}{\partial y} = x^3 + 4xy + 1$

Javob: $f_x = 3x^2y + 2y^2 - 3$, $f_y = x^3 + 4xy + 1$

Masala 4.2 ⭐⭐⭐

$f(x, y) = e^{xy} \sin(x + y)$ uchun gradientni toping.

Yechim

$\frac{\partial f}{\partial x} = ye^{xy}\sin(x+y) + e^{xy}\cos(x+y) = e^{xy}[y\sin(x+y) + \cos(x+y)]$

$\frac{\partial f}{\partial y} = xe^{xy}\sin(x+y) + e^{xy}\cos(x+y) = e^{xy}[x\sin(x+y) + \cos(x+y)]$

Javob: $\nabla f = e^{xy}[(y\sin(x+y) + \cos(x+y)), (x\sin(x+y) + \cos(x+y))]$

Masala 4.3 ⭐⭐⭐

$f(x, y, z) = x^2 + y^2 + z^2$ uchun $(1, 2, 2)$ nuqtada gradient yo'nalishi va kattaligini toping.

Yechim

$\nabla f = (2x, 2y, 2z)$

$(1, 2, 2)$ da: $\nabla f = (2, 4, 4)$

Kattlik: $|\nabla f| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

Yo'nalish (birlik vektor): $\hat{n} = \frac{1}{6}(2, 4, 4) = \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$

Javob: Kattlik = $6$, Yo'nalish = $\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$

Masala 4.4 ⭐⭐⭐⭐

Dron balandligi: $h(x, y, t) = 100 - 0.1(x^2 + y^2) + 5\sin(t)$

Dron $(x, y) = (10, 10)$ da $t = \frac{\pi}{2}$ vaqtda harakat qilmoqda.

a) Hozirgi balandlik b) X yo'nalishda tik balandlik o'zgarishi c) Vaqt bo'yicha balandlik o'zgarish tezligi

Yechim

a) Balandlik: $h = 100 - 0.1(100 + 100) + 5\sin(\pi/2) = 100 - 20 + 5 = 85 \text{ m}$

b) X bo'yicha qisman hosila: $\frac{\partial h}{\partial x} = -0.2x = -0.2(10) = -2 \text{ m/m}$

Har 1m X ga borganda balandlik 2m kamayadi.

c) Vaqt bo'yicha: $\frac{\partial h}{\partial t} = 5\cos(t) = 5\cos(\pi/2) = 0 \text{ m/s}$

Javob: a) 85 m, b) -2 m/m, c) 0 m/s

Masala 4.5 ⭐⭐⭐⭐

$f(x, y) = x^2 - 4xy + y^2$ uchun Hessian matritsasini toping va $(0, 0)$ nuqta turini aniqlang.

Yechim

Qisman hosilalar: $f_x = 2x - 4y, \quad f_y = -4x + 2y$

Ikkinchi tartibli: $f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = f_{yx} = -4$

Hessian: $H = \begin{bmatrix} 2 & -4 \\ -4 & 2 \end{bmatrix}$

Determinant: $\det(H) = 4 - 16 = -12 \lt 0$

Determinant manfiy → Egar nuqta (saddle point)

Javob: Hessian = $\begin{bmatrix} 2 & -4 \\ -4 & 2 \end{bmatrix}$, $(0,0)$ egar nuqta

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5-bo'lim: Ekstremal qiymatlar (4 ta masala)

Masala 5.1 ⭐⭐

$f(x) = x^3 - 6x^2 + 9x + 2$ ning lokal ekstremumlarini toping.

Yechim

Kritik nuqtalar: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) = 0$

$x = 1$ va $x = 3$

Ikkinchi hosila testi: $f''(x) = 6x - 12$

• $f''(1) = -6 \lt 0$ → lokal maksimum
• $f''(3) = 6 \gt 0$ → lokal minimum

Qiymatlar:

• $f(1) = 1 - 6 + 9 + 2 = 6$ (max)
• $f(3) = 27 - 54 + 27 + 2 = 2$ (min)

Javob: Lokal max: $(1, 6)$, Lokal min: $(3, 2)$

Masala 5.2 ⭐⭐⭐

Robot qo'li uzunliklari: $L_1 = 1$ m, $L_2 = 0.8$ m.

End effector $(x, y) = (1.2, 0.6)$ ga yetishi kerak. Inverse kinematika formulasi yordamida $\theta_1$ va $\theta_2$ ni toping.

Yechim

Inverse kinematika formulalari:

Masofa: $d = \sqrt{x^2 + y^2} = \sqrt{1.44 + 0.36} = 1.342$ m

Kosinus teoremasi:

$\cos\theta_2 = \frac{d^2 - L_1^2 - L_2^2}{2L_1L_2} = \frac{1.8 - 1 - 0.64}{1.6} = 0.1$

$\theta_2 = \arccos(0.1) = 84.3°$

Theta1 uchun:

$\theta_1 = \arctan\frac{y}{x} - \arctan\frac{L_2\sin\theta_2}{L_1 + L_2\cos\theta_2}$

$\theta_1 = 26.6° - 38.7° = -12.1°$ (yoki elbow-down konfiguratsiya)

Elbow-up yechim: $\theta_1 = 19.5°$, $\theta_2 = 39.5°$

Javob: Elbow-up: $\theta_1 = 19.5°$, $\theta_2 = 39.5°$

Masala 5.3 ⭐⭐⭐⭐

Raketa traektoriyasi optimizatsiyasi:

Yoqilg'i sarfi: $F(\theta, v) = v^2 + 100\theta^2$

Cheklov: Maqsad nuqtaga yetish — $x = 1000$ m

$x = v \cdot t \cdot \cos\theta, \quad t = 100s$

Lagrange usuli bilan minimal yoqilg'i sarfini toping.

Yechim

Cheklov: $g(\theta, v) = v \cdot 100 \cdot \cos\theta - 1000 = 0$

Lagrangian: $\mathcal{L} = v^2 + 100\theta^2 - \lambda(100v\cos\theta - 1000)$

Zaruriy shartlar:

$\frac{\partial \mathcal{L}}{\partial v} = 2v - 100\lambda\cos\theta = 0$

$\frac{\partial \mathcal{L}}{\partial \theta} = 200\theta + 100\lambda v\sin\theta = 0$

$\frac{\partial \mathcal{L}}{\partial \lambda} = 100v\cos\theta - 1000 = 0$

3-tenglamadan: $v\cos\theta = 10$

Agar $\theta = 0$ (optimal yo'nalish):

• $v = 10$ m/s
• $\lambda = \frac{2v}{100} = 0.2$

Tekshirish: 2-tenglama: $200(0) + 0 = 0$ ✓

Javob: Optimal: $\theta = 0deg$, $v = 10$ m/s, $F_{min} = 100$

Masala 5.4 ⭐⭐⭐⭐

$f(x, y) = x^2 + y^2 - 2x - 4y + 5$ ning $x^2 + y^2 \leq 9$ sohasidagi global min va max ni toping.

Yechim

Ichki kritik nuqtalar: $\nabla f = (2x - 2, 2y - 4) = (0, 0)$ $x = 1, y = 2$

$(1, 2)$ soha ichida? $1 + 4 = 5 \lt 9$ ✓

$f(1, 2) = 1 + 4 - 2 - 8 + 5 = 0$

Chegara ($x^2 + y^2 = 9$):

Parametrik: $x = 3\cos t$, $y = 3\sin t$

$g(t) = 9 - 6\cos t - 12\sin t + 5 = 14 - 6\cos t - 12\sin t$

$g'(t) = 6\sin t - 12\cos t = 0$ $\tan t = 2$ $t = \arctan(2) \approx 63.4deg$ va $t \approx 243.4deg$

$t = 63.4deg$: $(\frac{3}{\sqrt{5}}, \frac{6}{\sqrt{5}})$ $f = 14 - 6 \cdot \frac{1}{\sqrt{5}} - 12 \cdot \frac{2}{\sqrt{5}} = 14 - \frac{30}{\sqrt{5}} \approx 0.58$

$t = 243.4deg$: $(-\frac{3}{\sqrt{5}}, -\frac{6}{\sqrt{5}})$ $f = 14 + \frac{30}{\sqrt{5}} \approx 27.42$

Javob: Global min = $0$ da $(1, 2)$, Global max $\approx 27.4$ da $(-\frac{3}{\sqrt{5}}, -\frac{6}{\sqrt{5}})$

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6-bo'lim: Taylor qatori (3 ta masala)

Masala 6.1 ⭐⭐

$f(x) = e^x$ uchun $x = 0$ atrofida 4-tartibli Taylor polinomini yozing.

Yechim

$f(0) = 1$, $f'(0) = 1$, $f''(0) = 1$, $f'''(0) = 1$, $f^{(4)}(0) = 1$

$P_4(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$

$= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

Javob: $P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

Masala 6.2 ⭐⭐⭐

$f(x) = \sqrt{1 + x}$ uchun $x = 0$ da 3-tartibli Taylor qatorini yozing va $\sqrt{1.1}$ ni taxminan hisoblang.

Yechim

$f(x) = (1+x)^{1/2}$

$f(0) = 1$

$f'(x) = \frac{1}{2}(1+x)^{-1/2}$, $f'(0) = \frac{1}{2}$

$f''(x) = -\frac{1}{4}(1+x)^{-3/2}$, $f''(0) = -\frac{1}{4}$

$f'''(x) = \frac{3}{8}(1+x)^{-5/2}$, $f'''(0) = \frac{3}{8}$

$\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}$

$x = 0.1$ uchun: $\sqrt{1.1} \approx 1 + 0.05 - 0.00125 + 0.0000625 \approx 1.0488$

Haqiqiy qiymat: $\sqrt{1.1} \approx 1.04881$

Javob: $P_3(x) = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}$, $\sqrt{1.1} \approx 1.0488$

Masala 6.3 ⭐⭐⭐⭐

Kichik burchaklar taxmini: $\sin\theta \approx \theta$ va $\cos\theta \approx 1 - \frac{\theta^2}{2}$

Malyatnik davri formulasi: $T = 2\pi\sqrt{\frac{L}{g}}$

Katta amplituda uchun aniqroq formula: $T = 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16} + ...\right)$

$\theta_0 = 30deg$ uchun xatoni hisoblang.

Yechim

$\theta_0 = 30deg = \frac{\pi}{6} \approx 0.524$ rad

Oddiy formula: $T_0 = 2\pi\sqrt{\frac{L}{g}}$

To'g'rilangan formula: $T = T_0\left(1 + \frac{(0.524)^2}{16}\right) = T_0(1 + 0.0172) = 1.0172 T_0$

Xato: $\text{Xato} = \frac{T - T_0}{T} \approx 1.7\%$

Javob: Kichik burchak taxmini 30deg da ~1.7% xato beradi

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7-bo'lim: Amaliy masalalar (3 ta masala)

Masala 7.1 ⭐⭐⭐

Raketa massasi: $m(t) = m_0 - \dot{m}t$, bu yerda $\dot{m} = 100$ kg/s, $m_0 = 50000$ kg.

Dvigatel kuchi: $F = 1,500,000$ N

a) Tezlanishni vaqt funksiyasi sifatida toping b) $t = 100s$ dagi tezlanish c) Max tezlanish (yoqilg'i tugagunga qadar)

Yechim

a) Tezlanish: $a(t) = \frac{F}{m(t)} - g = \frac{1500000}{50000 - 100t} - 9.81$

b) $t = 100s$: $a(100) = \frac{1500000}{50000 - 10000} - 9.81 = \frac{1500000}{40000} - 9.81 = 37.5 - 9.81 = 27.7 \text{ m/s}^2$

c) Max tezlanish: Yoqilg'i massasi: $m_{fuel} = 40000$ kg (faraz) Yonish vaqti: $t_b = \frac{40000}{100} = 400s$

$a_{max} = a(400) = \frac{1500000}{50000 - 40000} - 9.81 = 150 - 9.81 = 140.2 \text{ m/s}^2 \approx 14.3g$

Javob: a) $a(t) = \frac{1500000}{50000-100t} - 9.81$, b) 27.7 m/s², c) 140.2 m/s²

Masala 7.2 ⭐⭐⭐⭐

Quadcopter boshqarish:

Motor tezligi $\omega$ va ko'tarish kuchi $T$ orasidagi bog'liqlik: $T = k\omega^2$

bu yerda $k = 1.5 \times 10^{-5}$ Ns²/rad².

a) $\frac{dT}{d\omega}$ ni toping b) $\omega = 800$ rad/s da sezuvchanlik c) Agar $\omega$ ±5% o'zgarsa, $T$ qancha o'zgaradi?

Yechim

a) Hosila: $\frac{dT}{d\omega} = 2k\omega = 3 \times 10^{-5} \omega$

b) $\omega = 800$ rad/s da: $\frac{dT}{d\omega}\bigg|_{800} = 3 \times 10^{-5} \times 800 = 0.024 \text{ N/(rad/s)}$

$T(800) = 1.5 \times 10^{-5} \times 640000 = 9.6$ N

c) 5% o'zgarish: $\Delta T \approx \frac{dT}{d\omega} \cdot \Delta\omega = 0.024 \times (0.05 \times 800) = 0.024 \times 40 = 0.96 \text{ N}$

Foizda: $\frac{0.96}{9.6} = 10\%$

Eslatma

$T \propto \omega^2$ bo'lgani uchun, 5% $\omega$ o'zgarishi ~10% $T$ o'zgarishiga olib keladi.

Javob: a) $\frac{dT}{d\omega} = 3 \times 10^{-5}\omega$, b) 0.024 N/(rad/s), c) ±10%

Masala 7.3 ⭐⭐⭐⭐

PID kontroller tuning:

Sistemaning xatosi: $e(t) = e^{-0.5t}\sin(2t)$

PID kontrol signali: $u(t) = K_p e(t) + K_d \frac{de}{dt}$

$K_p = 10$, $K_d = 2$ uchun $u(t)$ ni toping va $t = 1s$ da qiymatini hisoblang.

Yechim

Hosila: $\frac{de}{dt} = -0.5e^{-0.5t}\sin(2t) + e^{-0.5t} \cdot 2\cos(2t)$ $= e^{-0.5t}[2\cos(2t) - 0.5\sin(2t)]$

Kontrol signali: $u(t) = 10e^{-0.5t}\sin(2t) + 2e^{-0.5t}[2\cos(2t) - 0.5\sin(2t)]$ $= e^{-0.5t}[10\sin(2t) + 4\cos(2t) - \sin(2t)]$ $= e^{-0.5t}[9\sin(2t) + 4\cos(2t)]$

$t = 1s$ da: $u(1) = e^{-0.5}[9\sin(2) + 4\cos(2)]$ $= 0.6065[9 \times 0.909 - 4 \times 0.416]$ $= 0.6065[8.18 - 1.66]$ $= 0.6065 \times 6.52 = 3.95$

Javob: $u(t) = e^{-0.5t}[9\sin(2t) + 4\cos(2t)]$, $u(1) \approx 3.95$

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Qiyinlik darajasi bo'yicha

Daraja	Masalalar soni
⭐⭐ Oson	8
⭐⭐⭐ O'rta	13
⭐⭐⭐⭐ Murakkab	9
Jami	30