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a/b=c/d=>a/c=b/d(theo tính chất dãy tỉ số bằng nhau)
a/c=b/d=a-b/c-d=>(a-b/c-d)2014=a2014/c2014=b2014/d2014(theo t/c dãy tỉ số = nhau)
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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
\(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\\ \dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}\\ \left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{b^{2014}}{d^{2014}}\\ \RightarrowĐPCM\)
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
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) Theo t/c dãy tỷ số băng nhau ta có
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\left(\frac{a-b}{c-d}\right)^{2014}=\frac{a^{2014}}{c^{2014}}=\frac{b^{2014}}{d^{2014}}\) Theo t/c dãy tỷ số bằng nhau
\(\left(\frac{a-b}{c-d}\right)^{2014}=\frac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) (dpcm)
v~ cay wa đề mình là (a+b)^2014/(c+d)^2014= a^2014+b^2014/ c^2014+d^2014