Cho a, b, c khác nhau đôi một và 1/a+ 1/b+ 1/c =0. Rút gọn biểu thức:

Trả lời

Từ giả thiết suy ra ab + bc + ac = 0 nên

_{$a^2+2bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)$}

Tương tự: $b^2+2ac=\left(b-a\right)\left(b-c\right)$,

_{$c^2+2ab=\left(c-a\right)\left(c-b\right)$}

a) $\text{M=}\frac{\text{1}}{{\text{a}}^{\text{2}}\text{+2bc}}\text{+}\frac{\text{1}}{{\text{b}}^{\text{2}}\text{+2ac}}\text{+}\frac{\text{1}}{{\text{c}}^{\text{2}}\text{+2ab}}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0$

b) $\text{N=}\frac{\text{bc}}{{\text{a}}^{\text{2}}\text{+2bc}}\text{+}\frac{\text{ca}}{{\text{b}}^{\text{2}}\text{+2ac}}\text{+}\frac{\text{ab}}{{\text{c}}^{\text{2}}\text{+2ab}}=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}=1$

c) $\text{P=}\frac{{\text{a}}^{\text{2}}}{{\text{a}}^{\text{2}}\text{+2bc}}\text{+}\frac{{\text{b}}^{\text{2}}}{{\text{b}}^{\text{2}}\text{+2ac}}\text{+}\frac{{\text{c}}^{\text{2}}}{{\text{c}}^{\text{2}}\text{+2ab}}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}=1$.