Question: 9. Amallarni bajaring:
$\frac{2}{m} + \frac{5}{m-3n} - \frac{m-2n}{9n^2-m^2}$
$\frac{7m^2-18n^2+17mn}{n(m^2-9n^2)}$
1. Muammoni bayon qilamiz: quyidagi ifodani soddalashtiring:
$$\frac{2}{m} + \frac{5}{m-3n} - \frac{m-2n}{9n^2-m^2}$$
va
$$\frac{7m^2-18n^2+17mn}{n(m^2-9n^2)}$$
2. Birinchi ifodani soddalashtirish uchun umumiy maxrajni topamiz. E'tibor bering, $9n^2 - m^2$ ni farqlash formulasiga ko'ra:
$$9n^2 - m^2 = (3n - m)(3n + m)$$
3. Shuningdek, $m^2 - 9n^2 = (m - 3n)(m + 3n)$.
4. Endi birinchi ifodani yozamiz:
$$\frac{2}{m} + \frac{5}{m-3n} - \frac{m-2n}{(3n - m)(3n + m)}$$
5. $9n^2 - m^2 = -(m^2 - 9n^2)$, shuning uchun:
$$(3n - m) = -(m - 3n)$$
Shunday qilib, $\frac{m-2n}{9n^2 - m^2} = -\frac{m-2n}{(m-3n)(m+3n)}$
6. Umumiy maxraj $m(m-3n)(m+3n)$ bo'ladi.
7. Har bir kasrni umumiy maxrajga keltiramiz:
$$\frac{2}{m} = \frac{2(m-3n)(m+3n)}{m(m-3n)(m+3n)}$$
$$\frac{5}{m-3n} = \frac{5m(m+3n)}{m(m-3n)(m+3n)}$$
$$- \frac{m-2n}{(3n - m)(3n + m)} = \frac{m-2n}{(m-3n)(m+3n)} = -\frac{m-2n}{(m-3n)(m+3n)} = -\frac{m-2n}{(m-3n)(m+3n)}$$
Ammo bu ifoda $-\frac{m-2n}{(m-3n)(m+3n)}$ ga teng, shuning uchun umumiy maxrajga keltirilganda:
$$- \frac{m-2n}{(m-3n)(m+3n)} = - \frac{m-2n}{(m-3n)(m+3n)} \times \frac{m}{m} = - \frac{m(m-2n)}{m(m-3n)(m+3n)}$$
8. Endi barcha kasrlarni qo'shamiz:
$$\frac{2(m-3n)(m+3n) + 5m(m+3n) - m(m-2n)}{m(m-3n)(m+3n)}$$
9. Yuqoridagi ifodani ochamiz:
$$2(m^2 - 9n^2) + 5m^2 + 15mn - m^2 + 2mn = 2m^2 - 18n^2 + 5m^2 + 15mn - m^2 + 2mn$$
10. Soddalashtiramiz:
$$ (2m^2 + 5m^2 - m^2) + (-18n^2) + (15mn + 2mn) = 6m^2 - 18n^2 + 17mn$$
11. Shunday qilib, birinchi ifoda:
$$\frac{6m^2 - 18n^2 + 17mn}{m(m-3n)(m+3n)}$$
12. Ikkinchi ifodani ko'rib chiqamiz:
$$\frac{7m^2 - 18n^2 + 17mn}{n(m^2 - 9n^2)}$$
13. $m^2 - 9n^2 = (m-3n)(m+3n)$ ekanligini eslatamiz.
14. Shunday qilib, ikkinchi ifoda:
$$\frac{7m^2 - 18n^2 + 17mn}{n(m-3n)(m+3n)}$$
15. Natijalar:
Birinchi ifoda: $$\frac{6m^2 - 18n^2 + 17mn}{m(m-3n)(m+3n)}$$
Ikkinchi ifoda: $$\frac{7m^2 - 18n^2 + 17mn}{n(m-3n)(m+3n)}$$