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23 tháng 12 2018

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

22 tháng 12 2015

\(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\Leftrightarrow\frac{a+b}{ab}=-\frac{a+b}{c\left(a+b+c\right)}\)=> 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

 

13 tháng 3 2019

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

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)

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

\(P=0\)

27 tháng 2 2020

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

30 tháng 11 2017

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

12 tháng 3 2020

a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

12 tháng 3 2020

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999