Cho \(a+b+c=\frac{1}{2017}\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2017\). Tính \(\left(a^{2009}+b^{2009}+c^{2009}\right)\left(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\right)\)
Những câu hỏi liên quan
Bài 1: Tìm x biết: \(\left|x-\frac{2}{3}\right|-\left|x-7\right|=\frac{5}{3}\)
Bìa 2: Cho \(\frac{a}{b}=\frac{c}{d}\) và b+d\(\ne0\) . Chứng minh rằng \(\frac{a^{2009}+c^{2009}}{b^{2009}+d^{2009}}=\frac{\left(a+c\right)^{2009}}{\left(b+d\right)^{2009}}\)
cho a,b,c khac 0 ; a++b+c khac 0 thoa man \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
CMR\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{ca+cb+c^2+ab}{abc\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b\left(a+c\right)+c\left(a+c\right)\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow a+b=0\Rightarrow a=-b\Rightarrow a^{2009}=-b^{2009}\)
\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{c^{2009}}\) (1)
\(\frac{1}{a^{2009}+b^{2009}+c^{2009}}=\frac{1}{c^{2009}}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\) (đpcm)
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Cho \(\frac{a}{b}=\frac{c}{d}\) CMR : \(\frac{a^{2009}+c^{2009}}{b^{2009}+d^{2009}}=\frac{\left(a+c\right)^{2009}}{\left(b+d\right)^{2009}}\)
Ta có: a/b=c/d
Áp dụng t/c dãy tỉ số bằng nhau:
a/b=c/d=(a+c)/(b+d)
=>(a/b)2009=(c/d)2009=(a+c)2009/(b+d)2009(1)
a/b=c/d => (a/b)2009=(c/d)2009
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
(a/b)2009=(c/d)2009=a2009/b2009=c2009/d2009=(a2009+c2009)/(b2009+d2009)(2)
Từ (1)(2)=>....................
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cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
chứng minh \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=a^{2009}+b^{2009}+c^{2009}\)
P/s: Đề đúng phải là CM \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc-abc=0\)
\(\Leftrightarrow\left(a^2b+ab^2\right)+\left(c^2a+bc^2\right)+\left(ca^2+2abc+b^2c\right)=0\)
\(\Leftrightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+c^2+bc+ca\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> a+b=0 hoặc b+c=0 hoặc c+a=0
=> a=-b hoặc b=-c hoặc c=-a
Không mất tổng quát g/sử a=-b
Khi đó: \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=-\frac{1}{b^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{c^{2009}}\)
và \(\frac{1}{a^{2009}+b^{2009}+c^{2009}}=\frac{1}{-b^{2009}+b^{2009}+c^{2009}}=\frac{1}{c^{2009}}\)
=> \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)
nhưng đề lại ghi như trên
HELP ME:
Cho a+b+c = 2009 và\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2009}\)
Tính P = ( a - 2009 ).(b2 - 20092).(c3-20093)
Ta có :
\(a+b+c=2009\)
\(\Rightarrow\frac{1}{a+b+c}=\frac{1}{2009}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
\(\Rightarrow\frac{a+b}{ab}+\frac{\left(a+b+c\right)-c}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\left(a+b\right)\left(\frac{c^2+ab+bc+ca}{abc\left(a+b+c\right)}\right)=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}a+b=0\\b+c=0\\c+a=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}a=2009\\b=2009\\c=2009\end{array}\right.\)
(+) a = 2009
=> P = 0
(+) b = 2009
=> P = 0
(+) c = 2009
=> P = 0
Vậy P = 0
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Thu gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
\(B=\frac{\left(a+2008\right)!+\left(a+2009\right)!}{\left(a+2008\right)!-\left(a+2009!\right)}\)
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: \(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)
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)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a=-b\left(h\right)b=-c\left(h\right)c=-a\)
Thay vào tính nốt
cho a , b, c khác 0 và a+b+c khác 0 thoả 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 2 số đối nhau từ đó suy ra 1/(a^2009) + 1/(b^2009) + 1/(c^2009) = 1/(a^2009+b^2009+c^2009)
cho \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\) (với a,b,c,d khác 0 và b khác + d)
chứng minh rằng : \(\frac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\left(\frac{a}{b}\right)^{2009}\)
Vì \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^{2009}}{b^{2009}}=\frac{c^{2009}}{d^{2009}}=\left(\frac{a}{b}\right)^{2009}=\frac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}\)( áp dụng tc của dãy tỉ số bằng nhau )
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