Question

[minh nguyen thi]

Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) .CMR:

\(\dfrac{1}{a^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}\)

HELP ME !

Answer 1

[Phùng Khánh Linh]

Xuất phát từ giả thiết , ta có :

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

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

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

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

=> \(a\left(ab+bc+ac\right)+b\left(ab+bc+ac\right)+c\left(ab+bc+ac\right)-abc=0\)=> a²b + abc + a²c + ab² + b²c + abc + abc + bc² + ac² - abc = 0

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

=> ab( a + b + c) + ac( a + b + c ) + bc( b + c) = 0

=> ( a + b + c)a( b + c) + bc( b + c) = 0

=> ( b + c)( a² + ab + ac + bc) = 0

=> ( b + c)( a + b)( c + a) = 0

Suy ra :

* b = -c

*a = -b

* c = -a

TH1 :Với b = -c

\(VT=\dfrac{1}{a^{1995}}+\dfrac{1}{\left(-c\right)^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}}\)

\(VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{a^{1995}+\left(-c\right)^{1995}+c^{1995}}=\dfrac{1}{a^{1995}}=VT\)

TH2 : với a = -b

\(VT=\dfrac{1}{\left(-b\right)^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{c^{1995}}\)

\(VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{\left(-b\right)^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{c^{1995}}=VT\)

TH3 . c = -a , Tương tự

Vậy , đẳng thức được Chứng minh