a^2017+b^2017/c^2017+c^2017=a^2017-b^2017/c^2017-d^2017
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8 tháng 12 2018
Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^{2017}=\left(\frac{b}{d}\right)^{2017}=\frac{a^{2017}}{c^{2017}}=\frac{b^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{a^{2017}-b^{2017}}{c^{2017}-d^{2017}}\left(đpcm\right)\)
VN
1
25 tháng 1 2017
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\left(\frac{a}{c}\right)^{2017}=\left(\frac{b}{d}\right)^{2017}=\left(\frac{a-b}{c-d}\right)^{2017}\) (1)
\(\left(\frac{a}{c}\right)^{2017}=\left(\frac{b}{d}\right)^{2017}=\frac{a^{2017}}{c^{2017}}=\frac{b^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}\) (2)
Từ (1) và (2) => \(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\left(\frac{a-b}{c-d}\right)^{2017}\) ( đpcm )
VH
11 tháng 12 2019
Ta có:
b2=a.c c2=b.d
\(\Rightarrow\frac{b}{c}=\frac{a}{b}\) \(\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) (1)
\(\Rightarrow\hept{\begin{cases}\left(1\right)=\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}\\\left(1\right)=\frac{a+b-c}{b+c-d}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\end{cases}}\)
\(\Rightarrow\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
Vậy \(\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
11 tháng 12 2019
Ta có: \(b^2=a\cdot c\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
\(c^2=b\cdot d\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}\)(3)
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)(4)
Từ (3) và (4) \(\Rightarrow\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)(đpcm)
2 tháng 6 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{b^{2017}\cdot k^{2017}+d^{2017}\cdot k^{2017}}{b^{2017}+d^{2017}}=k^{2017}\)
\(\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\dfrac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)
Do đó: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)