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15. \( \left\{\begin{array}{l}\text { EKUB } a ; b)=1 \\ \frac{a+b}{a-b}=\frac{14}{9}\end{array}\right. \) sistemasidan foydalanib, \( a \cdot b \) ning qiymatini toping. \( \begin{array}{llll}\text { A) } 5 & \text { B) } 115 & \text { C) } 23 & \text { D) } 28\end{array} \) \( \begin{array}{llll}\text { 16. Radiusi } 4 \sqrt{3} \text { bo'lgan sharga ichki } \\ \text { chizilgan silindr hajmining eng katta }\end{array} \) qiymatini toping. \( \begin{array}{llll}\text { A) } 128 \pi & \text { B) } 324 \pi & \text { C) } 256 \pi & \text { D) } 216 \pi\end{array} \)

Ask by Reed Lindsey. in Uzbekistan
Mar 10,2025

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### 15-misol Berilgan shartlar: \[ \gcd(a,b)=1,\quad \frac{a+b}{a-b}=\frac{14}{9}. \] Natijada: \[ a=23,\quad b=5. \] \[ a \cdot b = 115. \] Javob: \(115\) (B varianti). --- ### 16-misol Berilgan: shar radiusi \(R=4\sqrt{3}\), ichiga chizilgan silindr hajmining eng katta qiymatini topishimiz kerak. Natijada: \[ V_{\max} = 256\pi. \] Javob: \(256\pi\) (C varianti).

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### 15-misol Berilgan shartlar: \[ \gcd(a,b)=1,\quad \frac{a+b}{a-b}=\frac{14}{9}. \] Avvalo, hosil qilamiz: \[ \frac{a+b}{a-b}=\frac{14}{9} \implies 9(a+b)=14(a-b). \] Bu tenglamani yechamiz: \[ 9a+9b=14a-14b, \] \[ 9a+9b-14a+14b=0, \] \[ -5a+23b=0 \quad \text{yoki} \quad 5a=23b. \] Shundan: \[ a=\frac{23}{5}b. \] Chunki \(a\) va \(b\) butun sonlar bo‘lishi va \(\gcd(a,b)=1\), \(b\) \(5\) ga bo‘linishi shart. Demak, \(b=5k\) bo‘lsa: \[ a=23k. \] Ikki sonning EKUBi: \[ \gcd(23k,5k)=k\gcd(23,5)=k, \] chunki \(23\) va \(5\) bir-biriga tub. Shartga ko‘ra \(\gcd(a,b)=1\) bo‘lishi uchun \(k=1\). Natijada: \[ a=23,\quad b=5. \] So‘ng \(a \cdot b\) ni hisoblaymiz: \[ 23\times 5=115. \] Javob: \(115\) (B varianti). --- ### 16-misol Berilgan: shar radiusi \[ R=4\sqrt{3}, \] ichiga chizilgan silindr hajmining eng katta qiymatini topishimiz kerak. Silindrni shar ichiga chizishda, silindrning radiusi \(r\) va balandligi \(h\) orasidagi munosabat: \[ h=2\sqrt{R^2-r^2}. \] Silindr hajmi: \[ V=\pi r^2h=2\pi r^2\sqrt{R^2-r^2}. \] Eng katta hajmni topish uchun optimal \(r\) ni aniqlash natijasi, isbotlanganidek: \[ r^2=\frac{2}{3}R^2\quad\text{va}\quad h=\frac{2R}{\sqrt{3}}. \] Shunday qilib, maksimal hajm: \[ V_{\max}=\pi \left(\frac{2}{3}R^2\right)\cdot\frac{2R}{\sqrt{3}}=\frac{4\pi R^3}{3\sqrt{3}}. \] Endi \(R=4\sqrt{3}\) ni qo‘yamiz: \[ R^3=(4\sqrt{3})^3=64\,(3\sqrt{3})=192\sqrt{3}. \] Shunda: \[ V_{\max}=\frac{4\pi (192\sqrt{3})}{3\sqrt{3}}=\frac{768\pi\sqrt{3}}{3\sqrt{3}}=256\pi. \] Javob: \(256\pi\) (C varianti).

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Ajablanarli, lekin bu operandni ko'rayotgan bo'lsangiz, ehtimoliy kompyuter dasturlari ham sizga bu muammolarni hal qilishda yordam beradi! Masalan, matematik simulyatorlar yordamida bir qator murakkab tizimlarni aniqlik bilan yechishingiz mumkin. Ular sizga jadvallar, grafiklar va boshqalar orqali matematikadan zavqlanish imkonini beradi. Agar tezda hisoblash kerak bo'lsa, oddiy formulalardan foydalanishni unutmang! Masalan, \( \frac{a+b}{a-b}=\frac{14}{9} \) tenglamasida, \( a \) va \( b \) ni ajratib olish yaxshidir, shunda ulardan \( a \cdot b \) ni chiqarganingizda aniqroq qiymatga ega bo'lasiz. Bunda ehtiyot bo'ling, ko'pincha oddiy bo'lgan muammolar ko'p xatolarga olib kelishi mumkin.

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