Murakkab tenglama va tengsizliklar

Bu mavzuda DTM nimani so'raydi

DTM imtihonlarida murakkab tenglama va tengsizliklar bo'yicha quyidagi savol turlari tushadi:

1. Yangi o'zgaruvchi kiritish: Kvadrat uchhadlar ko'paytmasi va bikvadrat tenglamalar.
2. Qaytma tenglamalar: $ax^4 + bx^3 + cx^2 + bx + a = 0$ ko'rinishidagi simmetrik ifodalar.
3. Kasr-ratsional tenglamalar: Maxraj nollaridan kelib chiqadigan chet ildizlar.
4. Oraliqlar usuli: Juft karrali nollarda ishora saqlanishi va ajralgan yechimlar.
5. Butun yechimlar tahlili: Berilgan sohadagi butun ildizlar soni va yig'indisi.

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Asosiy formulalar va usullar

1. Yangi o'zgaruvchi kiritish usullari

• Bikvadrat tenglama: $ax^4 + bx^2 + c = 0 \implies t = x^2$ ($t \ge 0$).
• Guruhlash usuli: $(x + a)(x + b)(x + c)(x + d) = C$ Agar $a + d = b + c$ bo'lsa, $(x^2 + (a+d)x + ad)(x^2 + (b+c)x + bc) = C$ shaklida $t = x^2 + (a+d)x$ almashtirish qilinadi.
• Qaytma tenglama: $ax^4 + bx^3 + cx^2 + bx + a = 0 \implies a\left(x^2 + \dfrac{1}{x^2}\right) + b\left(x + \dfrac{1}{x}\right) + c = 0$ Bunda $t = x + \dfrac{1}{x}$ va $x^2 + \dfrac{1}{x^2} = t^2 - 2$.

2. Oraliqlar usuli qoidalari

1. $\dfrac{P(x)}{Q(x)} \ge 0$ yoki $\dfrac{P(x)}{Q(x)} \le 0$ shakliga keltiriladi.
2. Surat nollari noqat'iy tengsizlikda bo'yaladi, maxraj nollari doimo ochiq qoladi ($Q(x) \ne 0$).
3. $(x - x_0)^{2k}$ juft karrali noldan o'tganda ishora saqlanadi, toq karralida almashadi.

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Yechilgan misollar

1-misol: Yangi o'zgaruvchi kiritish

$(x^2 - 4x + 1)(x^2 - 4x + 3) = 15$ tenglamaning haqiqiy ildizlari ko'paytmasini toping.

A) $-4$ B) $-2$ C) $2$ D) $4$

Yechimi:

1. $t = x^2 - 4x + 2$ deb olamiz: $(t - 1)(t + 1) = 15 \implies t^2 - 1 = 15 \implies t^2 = 16 \implies t = \pm 4$
2. $t = 4 \implies x^2 - 4x - 2 = 0$. $D = 16 + 8 = 24 > 0$. Viyet teoremasiga ko'ra $x_1 x_2 = -2$.
3. $t = -4 \implies x^2 - 4x + 6 = 0$. $D = 16 - 24 = -8 < 0$ (haqiqiy ildiz yo'q).

Tezroq yo'l: $u = x^2 - 4x$ desak, $(u+1)(u+3) = 15 \implies u^2 + 4u - 12 = 0 \implies u = 2$ yoki $u = -6$. $x^2 - 4x - 2 = 0 \implies x_1 x_2 = -2$.

Javob: B

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2-misol: Qaytma tenglama

$2x^4 + 3x^3 - 16x^2 + 3x + 2 = 0$ tenglamaning barcha haqiqiy ildizlari yig'indisini toping.

A) $-3$ B) $-1{,}5$ C) $1{,}5$ D) $3$

Yechimi:

1. $x = 0$ ildiz emas. Tenglamani $x^2$ ga bo'lamiz: $2\left(x^2 + \dfrac{1}{x^2}\right) + 3\left(x + \dfrac{1}{x}\right) - 16 = 0$
2. $t = x + \dfrac{1}{x}$ bo'lsa, $2(t^2 - 2) + 3t - 16 = 0 \implies 2t^2 + 3t - 20 = 0$.
3. Ildizlar: $t_1 = 2{,}5$ va $t_2 = -4$.
4. $x + \dfrac{1}{x} = 2{,}5 \implies 2x^2 - 5x + 2 = 0 \implies x_1 = 2, x_2 = 0{,}5$.
5. $x + \dfrac{1}{x} = -4 \implies x^2 + 4x + 1 = 0 \implies D = 12 > 0$, ildizlar yig'indisi $-4$.
6. Barcha 4 ta ildiz haqiqiy. Yig'indi: $2{,}5 + (-4) = -1{,}5$.

Javob: B

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3-misol: Uchinchi darajali tenglama

$x^3 - 7x + 6 = 0$ tenglama ildizlarining kvadratlari yig'indisini toping.

A) $10$ B) $12$ C) $14$ D) $16$

Yechimi:

1. Koeffitsiyentlar yig'indisi $1 - 7 + 6 = 0 \implies x_1 = 1$.
2. Ko'phadni bo'lamiz: $(x - 1)(x^2 + x - 6) = 0 \implies (x - 1)(x + 3)(x - 2) = 0$.
3. Ildizlar: $1, 2, -3$.
4. Kvadratlar yig'indisi: $1^2 + 2^2 + (-3)^2 = 14$.

Tezroq yo'l: Viyet formulasidan: $x_1 + x_2 + x_3 = 0$ va $x_1 x_2 + x_2 x_3 + x_3 x_1 = -7$. $x_1^2 + x_2^2 + x_3^2 = 0^2 - 2(-7) = 14$.

Javob: C

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4-misol: Juft karrali nolli kasr tengsizlik

$\dfrac{(x - 3)^2 (x + 2)}{(x - 1)(x - 5)} \le 0$ tengsizlikning $[0; 6]$ kesmadagi butun yechimlari yig'indisini toping.

A) $7$ B) $9$ C) $10$ D) $12$

Yechimi:

1. Surat nollari: $x = -2$, $x = 3$ (juft karrali). Maxraj nollari: $x = 1, x = 5$.
2. Oraliqlar ishorasi: $(-\infty; -2] \cup (1; 5)$.
3. $x = 3$ da ifoda $0 \le 0$ bo'lib, yechimga kiradi.
4. $[0; 6]$ kesmadagi butun yechimlar: $x \in \{2, 3, 4\}$.
5. Yig'indi: $2 + 3 + 4 = 9$.

Tezroq yo'l: $x = 1$ va $x = 5$ kirmaydi. $x \in \{2, 3, 4\}$ sonlarini to'g'ridan-to'g'ri tekshirish kifoya.

Javob: B

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5-misol: Chet ildizli kasr tenglama

$\dfrac{x}{x - 2} - \dfrac{1}{x + 1} = \dfrac{6}{x^2 - x - 2}$ tenglamani yeching.

A) $\{-2; 2\}$ B) $-2$ C) $2$ D) $\varnothing$

Yechimi:

1. Aniqlanish sohasi: $x \ne 2$ va $x \ne -1$.
2. Umumiy maxrajga keltiramiz: $x(x + 1) - (x - 2) = 6 \implies x^2 + 2 = 6 \implies x^2 = 4 \implies x = \pm 2$
3. $x = 2$ maxrajni nolga aylantiradi (chet ildiz). Yagona ildiz: $x = -2$.

Javob: B

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Keng tarqalgan tuzoqlar

[!WARNING] 1. Tengsizlikda maxrajni ko'paytirib yuborish Maxraj ishorasi noma'lum bo'lsa, uni o'ng tomonga ko'paytirib bo'lmaydi. Barcha hadlarni chap tomonga o'tkazib, umumiy maxrajga keltiring.

2. Juft karrali nollarda ishora almashtirish $(x - a)^{2k}$ ifodasi nolga teng bo'lgan nuqtadan o'tganda ishora o'zgarmaydi.

3. Maxraj nollarini yechimga qo'shib yuborish Noqat'iy tengsizlikda ham maxraj nollari hech qachon yechimga kirmaydi.

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Mashq

1. $(x^2 - 5x)^2 - 2(x^2 - 5x) - 24 = 0$ tenglamaning eng katta va eng kichik ildizlari ayirmasini toping.

Yechimi

$t^2 - 2t - 24 = 0 \implies t_1 = 6, t_2 = -4$. $x^2 - 5x - 6 = 0 \implies x = 6, -1$. $x^2 - 5x + 4 = 0 \implies x = 4, 1$. Ildizlar: $-1, 1, 4, 6$. Ayirma: $6 - (-1) = 7$. Javob: 7

2. $(x - 1)(x - 2)(x - 3)(x - 4) = 24$ tenglamaning haqiqiy ildizlari yig'indisini toping.

Yechimi

$(x^2 - 5x + 4)(x^2 - 5x + 6) = 24$. $t = x^2 - 5x + 5 \implies t^2 - 1 = 24 \implies t = \pm 5$. $x^2 - 5x = 0 \implies x = 0, 5$. Ikkinchi tenglamada $D < 0$. Yig'indi: 5. Javob: 5

3. $\dfrac{x^2 - 9}{x - 3} = 6$ tenglamaning ildizlari sonini toping.

Yechimi

$x \ne 3$. $x + 3 = 6 \implies x = 3$ (chet ildiz). Ildizlar soni: 0 ta. Javob: 0

4. $\dfrac{x^2 - 4x + 3}{x^2 - 4} \le 0$ tengsizlikning butun yechimlari yig'indisini toping.

Yechimi

Oraliqlar: $(-2; 1] \cup (2; 3]$. Butun yechimlar: $-1, 0, 1, 3$. Yig'indi: 3. Javob: 3

5. $\dfrac{(x - 4)^2 (x + 1)}{x - 2} \le 0$ tengsizlikning butun yechimlari nechta?

Yechimi

Yechim: $[-1; 2) \cup \{4\}$. Butun yechimlar: $-1, 0, 1, 4$, jami 4 ta. Javob: 4

6. $x^4 - 13x^2 + 36 = 0$ tenglamaning musbat ildizlari ko'paytmasini toping.

Yechimi

$(x^2 - 4)(x^2 - 9) = 0 \implies x = \pm 2, \pm 3$. Musbat ko'paytma: $2 \cdot 3 = 6$. Javob: 6

7. $\dfrac{x - 1}{x + 2} \ge 2$ tengsizlikning eng katta butun manfiy yechimini toping.

Yechimi

$\dfrac{x - 1 - 2(x + 2)}{x + 2} \ge 0 \implies \dfrac{x + 5}{x + 2} \le 0 \implies [-5; -2)$. Eng katta butun manfiy yechim: $-3$. Javob: -3

8. $x^3 - 3x^2 - 4x + 12 = 0$ tenglama ildizlari kvadratlari yig'indisini toping.

Yechimi

$(x - 3)(x^2 - 4) = 0 \implies x = 3, \pm 2$. Kvadratlar yig'indisi: $3^2 + 2^2 + (-2)^2 = 17$. Javob: 17

9. $\dfrac{x^2 + 2x - 8}{x^2 + 1} < 0$ tengsizlikning butun yechimlari sonini toping.

Yechimi

$x^2 + 1 > 0$. $x^2 + 2x - 8 < 0 \implies (-4; 2)$. Butun yechimlar: $-3, -2, -1, 0, 1$, jami 5 ta. Javob: 5

10. $x + \dfrac{1}{x} = 3$ bo'lsa, $x^3 + \dfrac{1}{x^3}$ ning qiymatini toping.

Yechimi

$x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right)^3 - 3\left(x + \dfrac{1}{x}\right) = 27 - 9 = 18$. Javob: 18