CMR : Nếu \(\dfrac{a}{b}=\dfrac{c}{d}th\text{ì}\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)
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TH
Thiên Hi
25 tháng 3 2018
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MV
1 tháng 11 2017
\(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\\ \dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}\\ \left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{b^{2014}}{d^{2014}}\\ \RightarrowĐPCM\)
31 tháng 3 2017
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
23 tháng 5 2017
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)
AT
23 tháng 5 2017
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:
\(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)
S
8 tháng 11 2017
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) => \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)
26 tháng 5 2022
Từ \(\dfrac{a}{d}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\left(\dfrac{a-b}{c-d}\right)^{2014}\left(1\right)\)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)
12 tháng 8 2017
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
12 tháng 8 2017
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
TT
13 tháng 10 2017
\(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}=\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}\) (1)
Từ (1) \(\Rightarrow\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}=\dfrac{a+2014+a-2014}{b+2015+b-2015}\)
\(=\dfrac{a+2014-\left(a-2014\right)}{b+2015-\left(b-2015\right)}=\dfrac{2a}{2b}=\dfrac{4028}{4030}=\dfrac{a}{b}=\dfrac{2014}{2015}\) (2)
Từ (2) : \(\dfrac{a}{b}=\dfrac{2014}{2015}\Rightarrow\dfrac{a}{2014}=\dfrac{b}{2015}\) ( đpcm )
TM
13 tháng 10 2017
Ta có: \(\dfrac{a+2014}{a-2014}=\dfrac{b+2015}{b-2015}\) ( \(a\ne\pm2014;b\ne\pm2015\))
\(\Rightarrow\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+2014}{b+2015}=\dfrac{a-2014}{b-2015}=\dfrac{\left(a+2014\right)+\left(a-2014\right)}{\left(b+2015\right)+\left(b-2015\right)}=\dfrac{\left(a+2014\right)-\left(a-2014\right)}{\left(b+2015\right)-\left(b-2015\right)}=\dfrac{a+2014+a-2014}{b+2015+b-2015}=\dfrac{a+2014-a+2014}{b+2015-b+2015}=\dfrac{2a}{2b}=\dfrac{4018}{2030}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{2014}{2015}\)
\(\Rightarrow\dfrac{a}{2014}=\dfrac{b}{2015}\) (đpcm)
