Sonning moduli, modulli tenglama va tengsizliklar

Bu nima va nega kerak

Piyoda qaysi tomonga yurmasin, bosib o'tgan yo'li musbat kattalikdir. Koordinatalar to'g'ri chizig'ida nuqtalar orasidagi masofani ifodalash uchun modul (mutlaq qiymat) tushunchasi qo'llaniladi.

Muhandislikda xatolik chegarasi ($|x - x_0| \le \varepsilon$), fizikada tebranishlar va og'ishlar aynan modulli ifodalar orqali yoziladi.

Asosiy g'oya

1. Modul ta'rifi va geometrik ma'nosi

Ta'rif. Haqiqiy son $a$ ning moduli deb quyidagiga aytiladi: $|a| = \begin{cases} a, & \text{agar } a \ge 0, \\ -a, & \text{agar } a < 0. \end{cases}$

• $|a|$ — koordinata boshidan $a$ nuqtagacha bo'lgan masofa.
• $|a - b|$ — $a$ va $b$ nuqtalar orasidagi masofa.
• Ixtiyoriy $a \in \mathbb{R}$ uchun $|a| \ge 0$.

2. Asosiy xossalar

1. $|-a| = |a|$ va $|a|^2 = a^2$.
2. $\sqrt{a^2} = |a|$.
3. $|a \cdot b| = |a| \cdot |b|$ va $\left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|} \; (b \neq 0)$.
4. $|a + b| \le |a| + |b|$ (uchburchak tengsizligi). Tenglik faqat $ab \ge 0$ da bajariladi.

3. Modulli tenglamalar

• $|f(x)| = c$ ($c > 0$): $f(x) = c$ yoki $f(x) = -c$. ($c < 0$ bo'lsa yechim yo'q).
• $|f(x)| = g(x)$: $\begin{cases} g(x) \ge 0, \\ f(x) = g(x) \quad \text{yoki} \quad f(x) = -g(x). \end{cases}$
• $|f(x)| = |g(x)|$: $f(x) = g(x)$ yoki $f(x) = -g(x)$.
• Oraliqlar usuli: Modul nollari orqali sonlar o'qi oraliqlarga bo'linadi va har bir oraliqda modullar ochiladi.

4. Modulli tengsizliklar ($c > 0$)

• $|f(x)| \le c \iff -c \le f(x) \le c$.
• $|f(x)| \ge c \iff f(x) \ge c \quad \text{yoki} \quad f(x) \le -c$.
• $|f(x)| \le |g(x)| \iff (f(x) - g(x))(f(x) + g(x)) \le 0$.

Yechilgan misollar

1-misol. Sodda tenglamalar

Tenglamalarni yeching: a) $|3x - 5| = 7$ b) $|x^2 - 4x| = 5$

Yechilishi:

a) $3x - 5 = 7 \implies 3x = 12 \implies x = 4$. $3x - 5 = -7 \implies 3x = -2 \implies x = -\dfrac{2}{3}$. Javob: $x_1 = 4, \; x_2 = -\dfrac{2}{3}$.

b) $x^2 - 4x = 5 \implies x^2 - 4x - 5 = 0 \implies x_1 = 5, \; x_2 = -1$. $x^2 - 4x = -5 \implies x^2 - 4x + 5 = 0 \implies D = -4 < 0$ (ildiz yo'q). Javob: $x_1 = -1, \; x_2 = 5$.

2-misol. O'ng tomoni o'zgaruvchili tenglama

$|2x - 1| = x + 4$

Yechilishi: Shart: $x + 4 \ge 0 \implies x \ge -4$.

1. $2x - 1 = x + 4 \implies x = 5 \ge -4$.
2. $2x - 1 = -(x + 4) \implies 3x = -3 \implies x = -1 \ge -4$. Ikkala ildiz ham shartni qanoatlantiradi.

Javob: $x_1 = -1, \; x_2 = 5$.

3-misol. Oraliqlar usuli

$|x + 2| + |x - 3| = 7$

Yechilishi: Nollar: $x = -2$ va $x = 3$.

1. $x < -2$: $-(x + 2) - (x - 3) = 7 \implies -2x + 1 = 7 \implies x = -3$ (oraliqqa kiradi).
2. $-2 \le x \le 3$: $(x + 2) - (x - 3) = 7 \implies 5 = 7$ (yechim yo'q).
3. $x > 3$: $(x + 2) + (x - 3) = 7 \implies 2x - 1 = 7 \implies x = 4$ (oraliqqa kiradi).

Javob: $x_1 = -3, \; x_2 = 4$.

4-misol. Modulli tengsizliklar

a) $|3x - 2| \le 7$ b) $|2x - 1| \ge |x + 2|$

Yechilishi: a) $-7 \le 3x - 2 \le 7 \implies -5 \le 3x \le 9 \implies -\dfrac{5}{3} \le x \le 3$.

b) Kvadratlar ayirmasi: $(2x - 1)^2 - (x + 2)^2 \ge 0 \implies (x - 3)(3x + 1) \ge 0$. Nollar: $-\dfrac{1}{3}$ va $3$. Yechim: $x \in \left(-\infty; -\dfrac{1}{3}\right] \cup [3; +\infty)$.

Javob: a) $\left[-\dfrac{5}{3}; 3\right]$; b) $\left(-\infty; -\dfrac{1}{3}\right] \cup [3; +\infty)$.

Mashq

1. $|4x - 7| = 9$ tenglamani yeching.
2. $|3x + 2| = -5$ tenglamaning ildizlarini toping.
3. $|x^2 - 3x| = 4$ tenglamaning haqiqiy ildizlari sonini aniqlang.
4. $|3x - 2| = 2x + 7$ tenglamaning ildizlari yig'indisini toping.
5. $|x - 2| = |2x + 1|$ tenglamani yeching.
6. $|x - 1| + |x - 4| = 5$ tenglamani yeching.
7. $|2x - 5| < 7$ tengsizlikning eng katta butun yechimini toping.
8. $|4x + 3| \ge 11$ tengsizlikni yeching.
9. $|x - 3| \le |x + 1|$ tengsizlikni yeching.
10. $\sqrt{x^2 - 6x + 9} = 2x - 9$ tenglamani yeching.

Javoblar va yechim yo'llari

1. $4x - 7 = 9 \implies x = 4$; $4x - 7 = -9 \implies x = -0{,}5$. Javob: $x_1 = -0{,}5, \; x_2 = 4$.

2. Modul manfiy songa teng bo'lolmaydi. Javob: Ildizga ega emas ($\emptyset$).

3. $x^2 - 3x - 4 = 0 \implies x = 4, x = -1$. $x^2 - 3x + 4 = 0 \implies D < 0$. Jami $2$ ta ildiz. Javob: $2$ ta.

4. Shart: $2x + 7 \ge 0$. $3x - 2 = 2x + 7 \implies x = 9$; $3x - 2 = -2x - 7 \implies x = -1$. Ikkisi ham o'rinli. Yig'indi: $9 + (-1) = 8$. Javob: $8$.

5. $x - 2 = 2x + 1 \implies x = -3$; $x - 2 = -2x - 1 \implies x = \dfrac{1}{3}$. Javob: $x_1 = -3, \; x_2 = \dfrac{1}{3}$.

6. Nollar $1$ va $4$. $x < 1 \implies 5 - 2x = 5 \implies x = 0$. $1 \le x \le 4 \implies 3 = 5$ ($\emptyset$). $x > 4 \implies 2x - 5 = 5 \implies x = 5$. Javob: $x_1 = 0, \; x_2 = 5$.

7. $-7 < 2x - 5 < 7 \implies -1 < x < 6$. Eng katta butun yechim: $5$. Javob: $5$.

8. $4x + 3 \ge 11 \implies x \ge 2$; $4x + 3 \le -11 \implies x \le -3{,}5$. Javob: $(-\infty; -3{,}5] \cup [2; +\infty)$.

9. $(x - 3)^2 \le (x + 1)^2 \implies -6x + 9 \le 2x + 1 \implies -8x \le -8 \implies x \ge 1$. Javob: $[1; +\infty)$.

10. $|x - 3| = 2x - 9$. Shart: $x \ge 4{,}5$. $x - 3 = 2x - 9 \implies x = 6 \ge 4{,}5$. $x - 3 = -2x + 9 \implies x = 4 < 4{,}5$ (chet ildiz). Javob: $x = 6$.

Keng tarqalgan xatolar

1. $|f(x)| = g(x)$ da $g(x) \ge 0$ shartini tekshirmaslik. ✗ $|x - 5| = 2x + 1$ tenglamada $x = -6$ ni ildiz deb olish. ✓ $x = -6$ da $2(-6) + 1 = -11 < 0$. Modul manfiy songa teng bo'la olmaydi.

2. Modulli tengsizlikni bir tomonlama ochish. ✗ $|2x - 1| \le 5 \implies 2x - 1 \le 5 \implies x \le 3$. ✓ Chap tomon $-5$ ni unutmaslik kerak: $-5 \le 2x - 1 \le 5 \implies -2 \le x \le 3$.

3. Manfiy bo'lishi mumkin bo'lgan tomonni kvadratga oshirish. ✗ Har qanday tengsizlikni to'g'ridan-to'g'ri kvadratga oshirish. ✓ Faqat ikkala tomoni ham qat'iy nomanfiy bo'lgan tengsizliklargina kvadratga oshiriladi.

4. Oraliqlar usulida oraliqqa tegishli bo'lmagan ildizni olish. ✗ $x < 1$ oraliqda yechib chiqqan $x = 3$ ni qabul qilish. ✓ Ildiz ko'rilayotgan oraliqqa tushmasa, u chet yechim hisoblanadi.

DTM uslubidagi test

Ushbu mavzu bo'yicha tayyorlangan 5 ta test savoli platformada interaktiv tarzda ishlaydi: o'z javobingizni tanlang va to'liq yechimni ko'ring.

Keyingi qadam

Modul qoidalarini mustahkamladik. Endi umumiy ildiz va daraja tushunchasini birlashtiruvchi n-darajali ildiz va ratsional ko'rsatkichli daraja mavzusiga o'tamiz.