Question

Cho A, B, C là 3 số thực khác 0 thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Tính giá trị biểu thức M = \(\left(a^{27}+b^{27}\right)\cdot\left(b^{41}+c^{41}\right)\cdot\left(c^{2017}+a^{2017}\right)\)

Answer 1

Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+abc+a^2c+b^2a+b^2c+abc+bc^2+ac^2=0\)

\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(ab+ac+bc+c^2\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left[a\left(b+c\right)+c\left(b+c\right)\right]\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)

Thay vào từng TH suy ra M=0