Question
1) \( \frac{3 b}{b-2}, \frac{4 b}{b+2}, \frac{12 b^{2}}{b^{2}-4} \) \( 5 x \quad 8 x \)
Ask by Wang Lang. in Uzbekistan
Jan 26,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\[
\frac{3b}{b-2}, \quad \frac{4b}{b+2}, \quad \frac{12b^{2}}{b^{2}-4}
\]
Bu ifodalarni soddalashtirish uchun:
1. **\(\frac{12b^{2}}{b^{2}-4}\)** ni faktorlash:
\[
\frac{12b^{2}}{(b-2)(b+2)}
\]
2. **\(\frac{3b}{b-2}\) va \(\frac{4b}{b+2}\)** ifodalarini qo'shish yoki ayirish mumkin emas, chunki ular turli paydalarga ega.
3. **5x va 8x** ifodalarini tahlil qilish uchun qo'shimcha ma'lumot kerak.
Xulosa: Bu ifodalarni soddalashtirish uchun qo'shimcha ma'lumot yoki vazifa kerak.
Solution
Quyidagi ifodalarni tahlil qilaylik va soddalashtiraylik:
### 1. Ifodalar bilan ishlash:
\[
\frac{3b}{b-2}, \quad \frac{4b}{b+2}, \quad \frac{12b^{2}}{b^{2}-4}
\]
**Soddalashtirish:**
- **Birlamchi ifoda:**
\[
\frac{12b^{2}}{b^{2}-4}
\]
Bu ifodani yanada soddalashtirish uchun, $b^{2}-4$ ni faktorlash mumkin:
\[
b^{2} - 4 = (b - 2)(b + 2)
\]
Shunday qilib:
\[
\frac{12b^{2}}{(b-2)(b+2)}
\]
- **Boshqa ikki ifoda:**
\[
\frac{3b}{b-2} \quad \text{va} \quad \frac{4b}{b+2}
\]
Bu ikki kiritilgan ifodalarni alohida tahlil qilish mumkin, lekin ular aloqador bo'lishi uchun qo'shilishi yoki ayrilishi mumkin emas, chunki ular turli paydalar ($b-2$ va $b+2$) ga ega.
### 2. Ikkinchi qator ifodalar:
\[
5x \quad 8x
\]
**Tahlil:**
Bu ikki ifoda oddiy chiziqli algebraik ifodalardir:
- $5x$ va $8x$ lineer tenglamalar bo'lib, ular $x$ o'zgaruvchisiga nisbatan chiziqli bog'lanishni ko'rsatadi.
- Agar bu ifodalardan biror tenglama yoki nisbat tuzish kerak bo'lsa, qo'shimcha ma'lumot talab qilinadi.
### Xulosa:
Agar bu ifodalar bilan bog'liq qo'shimcha vazifa yoki savol bo'lsa, iltimos, aniqroq ma'lumot bering. Shu bilan birga, yuqorida keltirilgan soddalashtirish va tahlillar sizga yordam berishini umid qilaman.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
To simplify the expressions, you can notice that \( \frac{12 b^{2}}{b^{2}-4} \) can be factored as \( \frac{12 b^{2}}{(b-2)(b+2)} \). Thus, finding a common denominator allows you to combine the fractions effectively, which is crucial for operations like addition or subtraction. Next, always keep an eye on the excluded values to prevent division by zero. In this case, both \( b = 2 \) and \( b = -2 \) will make the denominators zero, so those values need to be avoided when solving equations or simplifying expressions involving these fractions!
