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Lời giải:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow \frac{abc}{c(a+b)}=\frac{abc}{a(b+c)}=\frac{bca}{b(c+a)}\)
\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)
\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)
\(\Rightarrow a=b=c\) (do $a,b,c>0$)
$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Đặt : \(\dfrac{a}{b}=\dfrac{c}{d}=k\) (k khác 0)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Khi đó:
+)\(\left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\)
\(=\left(\dfrac{b.\left(k-1\right)}{d.\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}\) (1)
+)\(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\)
\(=\dfrac{b^{2014}.\left(k^{2014}+1\right)}{d^{2014}.\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\) (2)
Từ (1) và (2) suy ra
(đ.p.c.m)
Tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) có thể viết \(\dfrac{a}{c}=\dfrac{b}{d}\). Theo tính chất của dãy tỉ số bằng nhau ta có: \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\) hay nâng lên lũy thừa 2014:
\(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)
Áp dụng lần nữa tính chất của tỉ số bằng nhau sẽ được:
\(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\)\(\Rightarrow\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{bk^{2014}+b^{2014}}{dk^{2014}+d^{2014}}=\dfrac{b\left(k^{2014}+b^{2013}\right)}{d\left(k^{2014}+d^{2013}\right)}\)
2 cái này thấy nó ko giống nhau lắm:v
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có:+) \(\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\) (1)
+) \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}.k^{2014}+b^{2014}}{d^{2014}.k^{2014}+d^{2014}}\)
\(=\dfrac{b^{2014}.\left(k^{2014}+1\right)}{d^{2014}.\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) => đpcm
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Vì \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)
Mà \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}=\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\) (1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) (2)
Từ (1);(2) => \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\\ \dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}\\ \left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{b^{2014}}{d^{2014}}\\ \RightarrowĐPCM\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)
Xét \(VT=\left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}\left(1\right)\)
Xét \(VP=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{b^{2014}k^{2014}+b^{2014}}{d^{2014}k^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM
Từ \(\dfrac{ab}{2014}=\dfrac{1}{c}\Rightarrow abc=2014\) thay vào \(A\) ta có:
\(A=\dfrac{abc\cdot a}{ab+abc\cdot a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac\cdot ab}{ab\left(ac+c+1\right)}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac+c+1}{ac+c+1}=1\Rightarrow A=1\)