$$ \frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}=\frac{1}{x(x+1)(x+2)} $$ tenglikni ganoatlantiradigan $$ a, b, c $$ larni toping. (A) $$ a=0,5 ; b=-1 ; c=0,5 $$. B) $$ a=-0,5 ; b=-1 ; c=0,5 $$. C) $$ a=0,5 ; b=1 ; c=0,5 $$. D) $$ a=1 ; b=2 ; c=0,5 $$.

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Berilgan tenglik:
$$
\frac{a}{x} + \frac{b}{x+1} + \frac{c}{x+2} = \frac{1}{x(x+1)(x+2)}
$$
ekvivalent bo‘lishi uchun $$a$$, $$b$$ va $$c$$ ni topamiz.

Avvalo, tenglikni umumiy paydaga keltiramiz:
$$
a(x+1)(x+2) + b x(x+2) + c x(x+1) = 1
$$
Har bir bo‘luvchini kengaytiramiz:
$$
a(x^2 + 3x + 2) + b(x^2 + 2x) + c(x^2 + x) = 1
$$
Bu tenglamani yakka x ga nisbatan tenglashtiramiz:
$$
(a + b + c)x^2 + (3a + 2b + c)x + 2a = 1
$$
Bu tenglama har bir daraja bo‘yicha teng bo‘lishi uchun quyidagi shartlar bajarilishi kerak:
1. $$a + b + c = 0$$ (x² koeffitsienti uchun)
2. $$3a + 2b + c = 0$$ (x koeffitsienti uchun)
3. $$2a = 1$$ (erkin a’zolar uchun)

Uchinchisi orqali:
$$
2a = 1 \quad \Rightarrow \quad a = 0.5
$$
Birinchi tenglamaga $$a = 0.5$$ qo‘shamiz:
$$
0.5 + b + c = 0 \quad \Rightarrow \quad b + c = -0.5
$$
Ikkinchi tenglamaga ham $$a = 0.5$$ qo‘shamiz:
$$
3(0.5) + 2b + c = 0 \quad \Rightarrow \quad 1.5 + 2b + c = 0 \quad \Rightarrow \quad 2b + c = -1.5
$$
Endi ikki tenglamani yechamiz:
$$
\begin{cases}
b + c = -0.5 \
2b + c = -1.5
\end{cases}
$$
Ikkinchi tenglamadan birinchi tenglamani ayrib:
$$
(2b + c) - (b + c) = -1.5 - (-0.5) \quad \Rightarrow \quad b = -1
$$
Endi $$b = -1$$ ni birinchi tenglamaga qo‘ymiz:
$$
-1 + c = -0.5 \quad \Rightarrow \quad c = 0.5
$$

Demak, to‘g‘ri javob:
**(A) $$a=0.5 ; b=-1 ; c=0.5$$**.
$$ a = 0.5 $$, $$ b = -1 $$, $$ c = 0.5 $$.

\( \frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}=\frac{1}{x(x+1)(x+2)} \) tenglikni ganoatlantiradigan \( a, b, c \) larni toping. (A) \( a=0,5 ; b=-1 ; c=0,5 \). B) \( a=-0,5 ; b=-1 ; c=0,5 \). C) \( a=0,5 ; b=1 ; c=0,5 \). D) \( a=1 ; b=2 ; c=0,5 \).

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