Question

cho a+b+c=2023 và 1/a+1/b+1/c=1/2023. Tính M =1/a^2023+1/b^2023+1/c^2023

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2023}=\dfrac{1}{a+b+c}\)

\(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

\(\left(a+b\right)\left[\dfrac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Đến đây bạn thay vào nữa là được nhé