Question

Cho 4 số nguyên dương a;b;c;d thỏa mãn điều kiện a + c = 2b và c(b + d) = bd . Chứng minh rằng : \(\left(\dfrac{a+c}{b+d}\right)^8=\dfrac{a^8+c^8}{b^8+d^8}\)

Answer 1

Ta có:

\(c.\left(b+d\right)=2bd\)

\(\Rightarrow bc+cd=2bd\)

Lại có: \(a+c=2b\)

Lấy vế chia vế được: \(\dfrac{bc+cd}{a+c}=\dfrac{2bd}{2b}=d\)

\(\Rightarrow bc+cd=ad+cd\)

\(\Rightarrow bc=ad\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

* \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)

\(\Rightarrow\left(\dfrac{a+c}{b+d}\right)^8=\left(\dfrac{a}{b}\right)^8=\dfrac{a^8}{b^8}\left(1\right)\)

* \(\dfrac{a}{b}=\dfrac{c}{d}=\left(\dfrac{a}{b}\right)^8=\left(\dfrac{c}{d}\right)^8\)

\(\Rightarrow\dfrac{a^8}{b^8}=\dfrac{c^8}{d^8}=\dfrac{a^8+c^8}{b^8+d^8}\left(2\right)\)

Từ (1) và (2) suy ra:

\(\left(\dfrac{a+c}{b+d}\right)^8=\dfrac{a^8+c^8}{b^8+d^8}\left(đpcm\right)\)