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: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)

Từ gt => (a+c)(b-d)=(b+d)(a-c)

nên ab+bc-ad-cd=ab+ad-bc-cd   =>   2bc=2ad     => bc=ad

=> \(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}\)(theo t/c dãy tỉ số bằng nhau)

Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)

khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)

\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)

Ta có điều phải chứng minh

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c+a-c}{b+d+b-d}=\dfrac{2a}{2b}=\dfrac{a}{b}\left(1\right)\)

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c-a+c}{b+d-b+d}=\dfrac{2c}{2d}=\dfrac{c}{d}\left(2\right)\)

Từ (1) và (2) ta có:

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

Đặt:

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

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}=\dfrac{b^{2017}k^{2017}-d^{2017}k^{2017}}{b^{2017}-d^{2017}}=\dfrac{k^{2017}\left(b^{2017}-d^{2017}\right)}{b^{2017}-d^{2017}}=k^{2017}\left(3\right)\)

\(\Rightarrow\left(\dfrac{a}{b}\right)^{2017}=\dfrac{a^{2017}}{b^{2017}}=\dfrac{b^{2017}k^{2017}}{b^{2017}}=k^{2017}\left(4\right)\)

Từ (3) và (4) ta có:

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

\(\Rightarrowđpcm\)

ok 1 tick rui nhe

b. Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{2a}{2b}=\dfrac{2c}{2d}\)

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

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

Mà \(k=\dfrac{a}{b}\Rightarrow k^{2017}=\left(\dfrac{a}{b}\right)^{2017}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}=\left(\dfrac{a}{b}\right)^{2017}\)

\(a+b+c=2017\Rightarrow A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-b}+\dfrac{c}{a+b+c-a}\)

\(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow A< 2\left(1\right)\)

\(A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow A>1\left(2\right)\)

từ (1) và (2) \(\Rightarrow1< A< 2\)

vay A \(\notin Z\)