Question

[Bùi Hiền Thảo]

Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\)

Chứng minh rằng: \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)

Answer 1

[soyeon_Tiểubàng giải]

Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(1\right)\)

\(\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(đpcm\right)\)

Answer 2

[Nguyễn Huy Tú]

Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k,c=d.k\)

Ta có: 

\(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(b.k\right)^{2017}+\left(d.k\right)^{2017}}{b^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+d^{2017}.k^{2017}}{b^{2017}+d^{2017}}=\frac{k^{2017}.\left(b^{2017}+d^{2017}\right)}{b^{2017}+d^{2017}}=k^{2017}\) (1)

\(\left(\frac{a+c}{b+d}\right)^{2017}=\left(\frac{b.k+d.k}{b+d}\right)^{2017}=\left[\frac{k.\left(b+d\right)}{b+d}\right]^{2017}=k^{2017}\) (2)

Từ (1) và (2) suy ra \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)

Vậy \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)