Question

Найдите значениt выражения

Answer 1

Ответ:

-4

Объяснение:

$a - b + c = 0, => a + c = b$

$\frac{12abc}{a^3 - b^3 + c^3} = \frac{12abc}{(a^3 + c^3) - b^3} = \frac{12abc}{(a + c) \cdot (a^2 - ac + c^2) - b^3} = \\\\= \frac{12abc}{b \cdot (a^2 - ac + c^2) - b^3} = \frac{12abc}{b \cdot (a^2 - ac + c^2 - b^2)} = \frac{12ac}{a^2 - ac + c^2 - b^2} = \\\\= \frac{12ac}{a^2 - ac + c^2 - (a + c)^2} = \frac{12ac}{a^2 - ac + c^2 - (a^2 + 2ac+c^2)} = \\\\= \frac{12ac}{a^2 - ac + c^2 - a^2 - 2ac - c^2} = \frac{{12ac}}{-3ac} = \frac{12}{-3} = -4$

Answer 2

$a-b+c=0\ \ \Rightarrow \ \ b=a+c\ \ ,\ \ abc\ne 0\\\\\\\dfrac{12abc}{a^3-b^3+c^3}=\dfrac{12abc}{(a+c)(a^2-ac+c^2)-b^3}=\dfrac{12abc}{b\cdot (a^2-ac+c^2)-b^3}=\\\\\\=\dfrac{12ac}{(a^2-ac+c^2)-b^2}=\Big[\ a^2-ac+c^2=(a+c)^2-3ac\; \Big]=\\\\\\=\dfrac{12ac}{(a+c)^2-3ac-b^2}=\dfrac{12ac}{b^2-3ac-b^2}=\dfrac{12ac}{-3ac}=-\dfrac{12}{3}=-4$