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Ta có : 1/a+1/b+1/c=1/a+b+c
suy ra:(ab+bc+ac)/abc=1/a+b+c
=>(ab+ac+bc)(a+b+c)=abc
=>a2b+ab2+a2c+ac2+b2c+bc2+3abc=abc
(a+b)(b+c)(a+c)=0
=>a=-b hc b=-c hc a=-c
ta sẽ dễ dàng c/m được với 3 t/h trên thì 1/a1995+1/b1995+1/c1995=1/a1995+b1995+c1995
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\Leftrightarrow\frac{a+b}{ab}=-\frac{a+b}{c\left(a+b+c\right)}\)=> a = -b hoặc ab +c(a+b+c) =0
+ a = -b => thay => dpcm
+ab + c(a+b+c) =0 =>(a+c)(b+c) =0 => a =-c hoạc b =-c thay => dpcm
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)
\(P=0\)
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
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000