Question

[Thắm Nguyễn Thị]

Cho \(a,b,c\ne0\) và \(a+b+c\ne0\) thỏa mãn điều kiện \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

Chứng minh rằng trong 3 số a, b, c có hai số đối nhau. Từ đó suy ra rằng: \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)

Answer 1

[Trang Nguyễn]

T>a có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

=>\(\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

=> \(\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

=> \(ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)=abc\)

=> \(a^2b+ab^2+abc+abc+b^2c+bc^2+ca^2+abc+ac^2=abc\)

=> \(a^2b+ab^2+b^2c+bc^2+ca^2+ac^2+2abc=0\)

=> \(\left(a^2b+2abc+bc^2\right)+\left(ab^2+2abc+ac^2\right)+\left(b^2c-2abc+ca^2\right)=0\)

=> \(b\left(a+c\right)^2+a\left(b+c\right)^2+c\left(a-b\right)^2=0\)

=> \(\hept{\begin{cases}a+c=0\\b+c=0\\a-b=0\end{cases}\Rightarrow\hept{\begin{cases}a=-c\\b=-c\\a=b\end{cases}}}\)

=> trong 3 số a,b,c có  2 số đối nhau  ( đpcm)

Thay a=-c ,b = -c vào \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{\left(-c\right)^{2019}}+\frac{1}{\left(-c\right)^{2019}}+\frac{1}{c^{2019}}\)

                                                                                    \(=-\frac{1}{c^{2019}}\)(1)

\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{\left(-c\right)^{2019}+\left(-c\right)^{2019}+c^{2019}}=-\frac{1}{c^{2019}}\)  (2)

Từ (1),(2) => \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019}+c^{2019}}\)  (đpcm)

Answer 2

[coolkid]

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

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

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

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

\(\Rightarrow a=-b\left(h\right)b=-c\left(h\right)c=-a\)

Thay vào tính nốt