Question

Cho a,b,c,d là 4 số khác 0; biết \(\frac{a}{b}=\frac{c}{d}\).Chứng minh rằng \(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)

Answer 1

vì \(\frac{a}{b}\)=\(\frac{c}{d}\)=>\(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\) 

áp dụng tính chất dãy tỉ số bằng nhau

=> \(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\)= \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\)=\(\frac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}\)=\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)(diều phải chứng minh

Answer 2

Từ \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra a=bk

           c=dk

Ta có

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

Ta có

\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\frac{\left(bk-b\right)^{2017}}{\left(dk-d\right)^{2017}}=\frac{\left(b\left(k-1\right)\right)^{2017}}{\left(d\left(k-1\right)\right)^{2017}}=^{\frac{b^{2017}}{d^{2017}}}\)(2)

Từ (1) và (2)

Ta suy ra

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

Answer 3

từ gt: \(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra ad=bc

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

suy ra \(a^{2017}+b^{2017}.\left(c-d\right)^{2017}=c^{2017}+d^{2017}.\left(a-b\right)^{2017}\)

\(a^{2017}+b^{2017}.c^{2017}-b^{2017}.d^{2017}=c^{2017}+d^{2017}.a^{2017}-d^{2017}.b^{2017}\)

theo mình nghĩ là\(b^{2017}.c^{2017}=d^{2017}.a^{2017}\)

bc=da

Answer 4

Xin lỗi nha! Vì 3 bn giải đúng nên mk sẽ dùng 3 nick để k cho 3 bn. Thanks các bn!!!