#)Giải :

Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\left(1\right)\)

Lại có : \(\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\left(\frac{a}{c}\right)^{2019}=\left(\frac{b}{d}\right)^{2019}=\left(\frac{a+b}{c+d}\right)^{2019}=\frac{\left(a+b\right)^{2019}}{\left(c+d\right)^{2019}}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{\left(a+b\right)^{2019}}{\left(c+d\right)^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\left(đpcm\right)\)

ko bieets banj oi

                                                            Bài giải

* Từ \(\frac{a}{b}=\frac{c}{d}\text{ }\Rightarrow\text{ }\frac{a}{c}=\frac{b}{d}\text{ }\Rightarrow\text{ }\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\text{ ( * ) }\)

* Từ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\text{ }\Rightarrow\text{ }\frac{a^{2019}}{c^{2019}}=\frac{\left(a-b\right)^{2019}}{\left(c-d\right)^{2019}}\left(\text{**}\right)\)

* Từ \(\left(\text{*}\right),\left(\text{**}\right)\Rightarrow\text{ ĐPCM}\)

Cứu mình với 9:00 sáng nay mình nộp bài rùi

- Nếu \(a=c=0\Rightarrow\left(\frac{a-b}{c-d}\right)^{2019}=\left(\frac{b}{d}\right)^{2019}=\frac{b^{2019}}{d^{2019}}\)

\(\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}=\frac{-b^{2019}}{-d^{2019}}=\frac{b^{2019}}{d^{2019}}\Rightarrow\left(\frac{a-b}{c-d}\right)^{2019}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\)

- Nếu \(a;c\ne0\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{2a^{2019}}{2c^{2019}}=\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\left(\frac{a-c}{b-d}\right)^{2019}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\)

Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

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

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{\left(a-b\right)^{2019}}{\left(c-d\right)^{2019}}\left(1\right)\)

Có \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}\)

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

\(\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\frac{\left(a-b\right)^{2019}}{\left(c-d\right)^{2019}}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\Leftrightarrow c\left(a+b+c\right)\left(a+b\right)=-ab\left(a+b\right)\)

\(\Leftrightarrow\left(ac+bc+c^2\right)\left(a+b\right)+ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> a=-b hoặc b=-c hoặc c=-a

không mất tính tổng quát ,giả sử a=-b, ta có:

\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{-b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\left(1\right)\)

\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\left(2\right)\)

Từ  (1) và (2) => đpcm

Tương tự với 2 trường hợp còn lại ta cũng có đpcm

Lời giải:

\(a^3+b^3=c^3+d^3\)

$\Leftrightarrow (a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$

Mà $a+b=c+d$ nên $ab(a+b)=cd(c+d)$

Đến đây ta xét 2TH:

TH $a+b=c+d=0$ thì $a^{2019}+b^{2019}=c^{2019}+d^{2019}=0$ (đpcm)

TH $a+b=c+d\neq 0$ thì $ab=cd\Leftrightarrow \frac{a}{d}=\frac{c}{b}$

Đặt $\frac{a}{d}=\frac{c}{b}=t\Rightarrow a=dt; c=bt$

Khi đó:

$a+b=c+d$

$\Leftrightarrow dt+b=bt+d\Leftrightarrow (t-1)(d-b)=0$

Nếu $t-1=0\Rightarrow a=d; c=b$

$\Rightarrow a^{2019}=d^{2019}; b^{2019}=c^{2019}$

$\Rightarrow a^{2019}+b^{2019}=c^{2019}+d^{2019}$ (đpcm)

Nếu $d-b=0\Leftrightarrow b=d\Rightarrow a=c$

$\Rightarrow a^{2019}+b^{2019}=c^{2019}+d^{2019}$ (đpcm)

Vậy..........