Question
18. \( \frac{1-g^{2} x}{1+t g^{2} x} \leq \frac{1}{2} \) tengsizlikning \( [0 ; \pi] \) oraliqdagi eng katta va eng kichik yechimlari yig' indisini toping. A) \( \frac{\pi}{12} \) B) \( \frac{\pi}{2} \) C) \( \frac{2 \pi}{3} \) D) \( \pi \) 19. \( (x-y)^{2}+(x+y)^{2}=6 \quad \) bo'lsa, \( (3 x-y)^{2}+(x+3 y)^{2} \) qiymatini toping. A) 18 [2,2 ball] \( ] \) B) 30 C) 36 D) 60 20. \( \frac{32}{x^{2}-10 x+33} \) kasrning qabul qilishi mumkin bo'lgan barcha natural qiymatlari yig'indisini toping. A) 9 B) 4 C) 10 D) 7
Ask by Hobbs Deleon. in Uzbekistan
Feb 19,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( (3x - y)^2 + (x + 3y)^2 = 30 \).
Solution
19. \( (x-y)^{2}+(x+y)^{2}=6 \) bo'lsa, \( (3x-y)^{2}+(x+3y)^{2} \) qiymatini toping.
\( (x-y)^{2}+(x+y)^{2}=6 \) tengsizlikni qayta yozsak, biz quyidagilarni olamiz:
\( x^{2}-2xy+y^{2}+x^{2}+2xy+y^{2}=6 \)
\( 2x^{2}+2y^{2}=6 \)
\( x^{2}+y^{2}=3 \)
\( (3x-y)^{2}+(x+3y)^{2} \) qiymatini toping:
\( (3x-y)^{2}+(x+3y)^{2} = 9x^{2}-6xy+y^{2}+x^{2}+6xy+9y^{2} \)
\( = 10x^{2}+10y^{2} \)
\( = 10(x^{2}+y^{2}) \)
\( = 10(3) \)
\( = 30 \)
Shuning uchun, \( (3x-y)^{2}+(x+3y)^{2} = 30 \).
Javob: B) 30
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Keling, birinchi masala haqida gaplashamiz. Tengsizlikni yechishda \( g(x) \) va \( t(x) \) kabi funksiyalarning tabiati va xossalarini o'rganish ulkan ahamiyatga ega. Har bir yechimni topishdan oldin, ularning oraliqlarda qanday xulq-atvor kasb etishini bilishimiz muhimdir. Keyingi masala esa algebraga oid. Siz \( (3x-y)^2 + (x+3y)^2 \) ifodasini almashtirish uchun berilgan tenglamadan foydalanishingiz mumkin. Ikkita yangi o'zgarish kiritish orqali masalani ancha osonlashtirishingiz mumkin, shuningdek, yig'indining qanchalik qiziqarli ko'rinishga ega bo'lishidan hayratda qoling!
