Question

[MIGHFHF]

Cho a;b;c khác 0 và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Chứng minh : \(\dfrac{1}{a^n} +\dfrac{1}{b^n}+\dfrac{1}{c^n}+\dfrac{1}{a^n+b^n+c^n}\)

Answer 1

[Trần Trung Nguyên]

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\Leftrightarrow a^2b+ab^2+abc+ac^2+abc+ac^2+abc+b^2c+bc^2=abc\Leftrightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc=0\Leftrightarrow\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(b^2c+abc\right)+\left(ac^2+bc^2\right)=0\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

TH1:a=-b

\(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n}-\dfrac{1}{a^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)(vì n lẻ)

\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{a^n-a^n+c^n}=\dfrac{1}{c^n}\)(vì n lẻ)

Suy ra \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)

Chứng minh tương tự trong các trường hợp b=-c và c=-a

Vậy \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}\)

Answer 2

[Trần Trung Nguyên]

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