Question

Cho a,b,c thoa man \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Tinh GT cua bieu thuc A=\(\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)\)

Answer 1

Lời giải:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)

\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)

\(\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)

\(\Rightarrow (a+b)(c+a)(c+b)=0\)

Do đó:

\(A=(a^3+b^3)(b^3+c^3)(c^3+a^3)\)

\(=(a+b)(a^2-ab+b^2)(b+c)(b^2-bc+c^2)(c+a)(c^2-ca+a^2)\)

\(=(a+b)(c+a)(c+b)[(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)]=0\)