Darslar 15 · 1 ta dars · Sh.A. Alimov, A.R. Xalmuxamedov, M.A. Mirzaxmedov. Algebra: umumiy o‘rta ta’lim maktablarining 9-sinfi uchun darslik. Qayta ishlangan 4-nashri. Toshkent: «O‘qituvchi» NMIU, 2019
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Tenglamalar sistemasini yechishning turli usullari
hadma-had qo‘shish · term-by-term additionsimmetrik sistema · symmetric systemyangi noma’lum kiritish · introducing a new unknowntenglamani tenglamaga bo‘lish · dividing one equation by another
Tushuntirish
Sistemani yechishning bir necha usuli bor va tenglamalar ko‘rinishiga qarab tanlanadi. 1) Tenglamalarni hadma-had qo‘shish yoki ayirish: x + y va xy kabi ifodalarni alohida topish mumkin bo‘ladi. 2) Ifodani almashtirish: agar y² yoki xy bitta tenglamadan x orqali ifodalansa, uni ikkinchi tenglamaga qo‘yamiz. 3) Simmetrik sistemalar (x va y ni almashtirganda o‘zgarmaydigan): s = x + y, p = xy deb olamiz; x² + y² = s² − 2p, chunki (x + y)² = x² + 2xy + y². So‘ng z² − sz + p = 0 ni yechamiz. 4) Bir tenglamani ikkinchisiga bo‘lish (x ≠ 0, y ≠ 0 ekanini tekshirib), masalan x²y = 4 va xy² = 2 dan x : y = 2. Har gal topilgan juftliklarni tekshiring, ayniqsa ildizli yoki kasrli tenglamalarda.
Namunalar
x² + xy = 15, y² + xy = 10: qo‘shsak (x + y)² = 25, x + y = ±5; ayirsak x² − y² = 5, ya’ni (x − y)(x + y) = 5, x − y = ±1. x + y = 5, x − y = 1 dan (3; 2); x + y = −5, x − y = −1 dan (−3; −2).
x + y² = 5, xy² = 4: y² = 5 − x, x(5 − x) = 4, x² − 5x + 4 = 0, x₁ = 1, x₂ = 4. x = 1 da y² = 4, y = ±2; x = 4 da y² = 1, y = ±1. Javob: (1; 2), (1; −2), (4; 1), (4; −1).
Sinfda faoliyat
«Usulni tanla»: o‘qituvchi sistemalar yozilgan kartochkalar tarqatadi; guruhlar avval qaysi usul qulay ekanini aytadi (qo‘shish, almashtirish, s va p, bo‘lish), so‘ng yechadi va javoblarni tekshiradi.
Mashq
1
x² + y² = 34, xy = 15 sistemani yeching.
(3; 5), (5; 3), (−3; −5), (−5; −3)
2
x²y = 4, xy² = 2 sistemani yeching.
(2; 1)
3
1/x + 1/y = 3/4, xy = 8 sistemani yeching.
(2; 4), (4; 2)
4
Tenglamani tenglamaga bo‘lishdan oldin nimani tekshirish kerak va nima uchun?
x ≠ 0 va y ≠ 0 ekanini: nolga bo‘lib bo‘lmaydi, bu holda yechimlar yo‘qolishi mumkin.