Cho biết: $\frac{a^{2}}{a+b}$$+$$\frac{b^{2}}{b+c}$$+$$\frac{c^{2}}{c+a}$$=2012$ Tính giá trị của biểu thức: $\frac{b^{2}}{a+b}$$+$$\frac{c^{2}}{b

Đáp án:

$\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}=2012$

Giải thích các bước giải:

Ta có:

$\quad \dfrac{a^2}{a+b} +\dfrac{b^2}{b+c} +\dfrac{c^2}{c+a}=2012$

$\to \dfrac{a^2- b^2 + b^2}{a+b} +\dfrac{b^2- c^2 + c^2}{b+c} +\dfrac{c^2- a^2 + a^2}{c+a}=2012$

$\to \dfrac{a^2 – b^2}{a+b} +\dfrac{b^2}{a+b}+\dfrac{b^2-c^2}{b+c} +\dfrac{c^2}{b+c}+\dfrac{c^2-a^2}{c+a}+\dfrac{a^2}{c+a}=2012$

$\to \dfrac{(a-b)(a+b)}{a+b} +\dfrac{(b-c)(b+c)}{b+c} +\dfrac{(c-a)(c+a)}{c+a} + \left(\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}\right)=2012$

$\to a-b + b-c + c – a + \left(\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}\right)=2012$

$\to \dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}=2012$