$$ \{a\}-a $$ ning kasr qismi bo'lsa, $$ |\{2 x\}|=0,1 $$ tenglama $$ [-3 ; 3] $$ oraliqda nechta yechimga ega? $$ x|x+2|-3|x+2|-x+3 \geq 0 $$ tengsizlikning eng kichik musbat butun va eng katta manfiy butun yechimlari ko'paytmasini toping.

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**1. Tenglama $$|\{2x\}| = 0.1$$ ni yechish:**

Bu tenglama $$\{2x\}$$ — $$2x$$ ning kasr qismi bo'lgani uchun $$\{2x\} = 0.1$$ bo'ladi. Buni yechish uchun quyidagi qadamlarni amalga oshiramiz:

$$
\{2x\} = 0.1 \implies 2x = k + 0.1 \quad \text{(bu yerda $$k$$ — butun son)}
$$
$$
x = \frac{k + 0.1}{2}
$$

Berilgan oraliq $$[-3; 3]$$ bo'lgani uchun:

$$
-3 \leq x \leq 3 \implies -6 \leq k + 0.1 \leq 6 \implies -6.1 \leq k \leq 5.9
$$

Bu shartga mos keluvchi $$k$$ butun sonlar $$-6$$ dan $$5$$ gacha bo'ladi. Demak, $$k$$ ning qiymatlari: $$-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5$$. Bu jami **12 ta** yechim beradi.

**2. Tengsizlik $$x|x+2| - 3|x+2| - x + 3 \geq 0$$ ni yechish:**

Tengsizlikni yechish uchun quyidagi holatlarni ko'rib chiqamiz:

**a) $$x + 2 \geq 0$$ (ya'ni, $$x \geq -2$$):**

$$
|x+2| = x + 2
$$
$$
x(x + 2) - 3(x + 2) - x + 3 \geq 0
$$
$$
x^2 + 2x - 3x - 6 - x + 3 \geq 0
$$
$$
x^2 - 2x - 3 \geq 0
$$
Kvadrat tenglamaning ildizlari: $$x = 3$$ va $$x = -1$$.

Shuning uchun:
$$
x \leq -1 \quad \text{yoki} \quad x \geq 3
$$
Ammo, bu holatda $$x \geq -2$$ bo'lganligi sababli, yechimlar:
$$
-2 \leq x \leq -1 \quad \text{yoki} \quad x \geq 3
$$

**b) $$x + 2 < 0$$ (ya'ni, $$x < -2$$):**

$$
|x+2| = -(x + 2)
$$
$$
x(-x - 2) - 3(-x - 2) - x + 3 \geq 0
$$
$$
- x^2 - 2x + 3x + 6 - x + 3 \geq 0
$$
$$
- x^2 + 0x + 9 \geq 0 \implies x^2 \leq 9
$$
$$
-3 \leq x \leq 3
$$
Ammo, bu holatda $$x < -2$$, demak yechim:
$$
-3 \leq x < -2
$$

**Umumiy yechim:**
$$
-3 \leq x \leq -1 \quad \text{yoki} \quad x \geq 3
$$

**Butun yechimlar:**

- **Eng katta manfiy butun yechim:** $$-1$$
- **Eng kichik musbat butun yechim:** $$3$$

**Ko'paytma:**
$$
-1 \times 3 = -3
$$

**Javob:**

1. Tenglama $$|\{2x\}| = 0.1$$ oraliqda **12 ta** yechimga ega.
2. Tengsizlikning eng kichik musbat butun yechimi **3** va eng katta manfiy butun yechimi **-1** bo'lib, ularning ko'paytmasi **-3** ga teng.
**1. Tenglama $$|\{2x\}| = 0.1$$ oraliqda 12 ta yechimga ega.** **2. Tengsizlikning eng kichik musbat butun yechimi 3 va eng katta manfiy butun yechimi -1. Ko'paytmasi -3 ga teng.**

\( \{a\}-a \) ning kasr qismi bo'lsa, \( |\{2 x\}|=0,1 \) tenglama \( [-3 ; 3] \) oraliqda nechta yechimga ega? \( x|x+2|-3|x+2|-x+3 \geq 0 \) tengsizlikning eng kichik musbat butun va eng katta manfiy butun yechimlari ko'paytmasini toping.

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