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Modul qatnashgan tenglamalar
maktab darsligi: algebra 8–9-sinf, son moduli, modulli tenglamalar
Maqsad: Modulli tenglamani modulni ochib, ikki holga ajratib yechish.
Qoida va g‘oya
|a| — a sonining son o‘qidagi 0 gacha masofasi, u hech qachon manfiy emas. |x| = c tenglamada: c > 0 bo‘lsa x = c yoki x = −c; c = 0 bo‘lsa x = 0; c < 0 bo‘lsa yechim yo‘q. |f(x)| = c da f(x) = c yoki f(x) = −c deb ikkita tenglama yechiladi. |f(x)| = |g(x)| da f(x) = g(x) yoki f(x) = −g(x). Masalan, |x − 3| = 5 dan x − 3 = 5 yoki x − 3 = −5, ya’ni x = 8 va x = −2. Javoblarni asl tenglamaga qo‘yib tekshirish foydali.
Namunaviy masalalar
|x − 3| = 5 ni yeching.
Yechim: x − 3 = 5 yoki x − 3 = −5. x = 8 yoki x = −2.
Javob: 8 va −2
Yechim: x − 3 = 5 yoki x − 3 = −5. x = 8 yoki x = −2.
Javob: 8 va −2
|2x − 1| = |x + 4| ni yeching.
Yechim: 2x − 1 = x + 4 dan x = 5. 2x − 1 = −x − 4 dan 3x = −3, x = −1.
Javob: 5 va −1
Yechim: 2x − 1 = x + 4 dan x = 5. 2x − 1 = −x − 4 dan 3x = −3, x = −1.
Javob: 5 va −1
|x + 2| = −3 ni yeching.
Yechim: Modul manfiy bo‘lmaydi, shuning uchun yechim yo‘q.
Javob: yechim yo‘q
Yechim: Modul manfiy bo‘lmaydi, shuning uchun yechim yo‘q.
Javob: yechim yo‘q
Tipik xatolar
- Modul tenglamasida faqat bitta ishorani olish: ikkita holni yozish kerak.
- |x| = −3 ga yechim topishga urinish.
- |f| = |g| da ikkinchi holni (f = −g) unutish.
Mashq
1
|x − 4| = 6 ni yeching. Ildizlar yig‘indisini toping.
8
2
|3x − 1| = 8 ni yeching.
3 va −7/3
3
|x| = 0 tenglamaning nechta yechimi bor?
1
4
|x + 1| = |x − 3| ni yeching.
1