Tính giá trị của phân thức C = (a^3 - b^3 + c^3 + 3abc)/((a + b)^2 + (b + c)^2 + (c - a)^2)

* Hướng dẫn giải

$$\begin{gathered} Ta có: a^3 - b^3 + c^3 + 3abc \\ = (a^3 + c^3 + 3a^{2}c + 3a{c}^2) - 3a^{2}c - 3a{c}^2 + 3abc - b^3 \\ = {(a + c)}^3 - b^3 - 3ac(a + c - b) \\ = (a + c - b)[{(a + c)}^2 + b(a + c) + b^2] - 3ac(a + c - b) \\ = (a + c - b)(a^2 + b^2 + c^2 + ab + bc - ac) \\ {(a + b)}^2 + {(b + c)}^2 + {(c - a)}^2 \\ = (a^2 + 2ab + b^2) + (b^2 + 2bc + c^2) + (c^2 - 2ac + a^2) \\ = 2a^2 + 2b^2 + 2c^2 + 2ab + 2bc - 2ac \\ = 2 (a^2 + b^2 + c^2 + ab + bc - ac) \end{gathered}$$

$=> C =\frac{\text{(a}+\text{c}-{\text{b)(a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+\text{ab}+\text{bc}-\text{ac)}}{{\text{2(a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+\text{ab}+\text{bc}-\text{ac)}}=\frac{\text{a}+\text{c}-\text{b}}{\text{2}}$

Mà a + c - b = 10 nên $C = \frac{\text{a}+\text{c}-\text{b}}{\text{2}}=\frac{\text{10}}{\text{2}}=\text{5}$

Đáp án D