Cho \(\frac{a}{b}=\frac{c}{d}\) CMR : \(\frac{a^{2009}+c^{2009}}{b^{2009}+d^{2009}}=\frac{\left(a+c\right)^{2009}}{\left(b+d\right)^{2009}}\)
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 \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\) (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 )
Vậy ...
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cho \(\frac{q+c}{b+d}\frac{a-c}{b-d}\) ( với a,b,c khác 0 và b khác cộng trừ d)
Chứng minh rằng: \(\frac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\left(\frac{a}{b}\right)^{2009}\)
Cho \(a+b+c=\frac{1}{2017}\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2017\). Tính \(\left(a^{2009}+b^{2009}+c^{2009}\right)\left(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\right)\)
Có: a+b+c0Leftrightarrowleft(a+b+cright)^20Leftrightarrow a^2+b^2+c^2+2left(ab+bc+acright)0 Leftrightarrow2009+2left(ab+bc+acright)0 Leftrightarrow ab+bc+ac-frac{2009}{2} Leftrightarrowleft(ab+bc+acright)^2a^2b^2+b^2c^2+a^2c^2+2abcleft(a+b+cright)left(-frac{2009}{2}right)^2 Leftrightarrowleft(ab+bc+acright)^2a^2b^2+b^2c^2+a^2c^2left(-frac{2009}{2}right)^2 Mặt khác: left(a^2+b^2+c^2right)^2a^4+b^4+c^4+2left(a^2b^2+b^2c^2+c^2a^2right) a^4+b^4+c^4+2.left(-frac{2009}{2}right)^22009^2Leftrightarrow...
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Có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow2009+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{2009}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\left(-\frac{2009}{2}\right)^2\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2=\left(-\frac{2009}{2}\right)^2\)
Mặt khác: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=a^4+b^4+c^4+2.\left(-\frac{2009}{2}\right)^2=2009^2\)
\(\Leftrightarrow a^4+b^4+c^4=2009^2-2.\left(-\frac{2009}{2}\right)^2=2009^2-2.\frac{2009^2}{2^2}=2009^2-\frac{2009^2}{2}\)
--Hà Phương--
Cho \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\) (Với a, b,c,d khác 0 và b khác d ,-d)
CMR \(\dfrac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\left(\dfrac{a}{b}\right)^{2009}\)
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
\(\Leftrightarrow ab-ad+cb-cd=ab+ad-cb-cd\)
=>-2ad=-2cb
=>ad=cb
=>a/b=c/d
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\dfrac{b^{2009}k^{2009}-d^{2009}k^{2009}}{b^{2009}-d^{2009}}=k^{2009}\)
\(\left(\dfrac{a}{b}\right)^{2009}=\left(\dfrac{bk}{b}\right)^{2009}=k^{2009}\)
Do đó: \(\dfrac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\left(\dfrac{a}{b}\right)^{2009}\)
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Cho dãy tỉ số bằng nhau : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}.CMR:\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
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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 dãy tỉ số bằng nhau : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}.CMR:\frac{\left(a-c\right)^2}{4}=\left(a-c\right).\left(b-c\right)\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
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Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
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