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

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

\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-\left(a+b+c\right)}{ac+bc+c^2}\)

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

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

\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)

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

=> a = - b hoặc b = - c hoặc a = - c

Xét a = - b ta có :

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

\(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{\left(-b^{2017}+b^{2017}\right)+c^{2017}}=\frac{1}{c^{2017}}\) (2)

Từ (1) ; (2) => \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)

Tới đây bạn xét tiếp 2 TH b = - c và c = - a nữa ta có đpcm nha

b ) TQ :

Nếu a +b +c khác 0; a;b;c khác 0 ; \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) thì \(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)

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

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

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

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{ac+bc+c^2}\)

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

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

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

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

=> a = - b hoặc b = - c hoặc c = - a 

Xét a = - b ta có \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{-b^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{c^{2017}}\)(1)

\(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{\left(-b^{2017}+b^{2017}\right)+c^{2017}}=\frac{1}{c^{2017}}\)(2)

Từ (1);(2) \(\Rightarrow\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)

Xét tiếp 2 TH b = - c hoặc c = - a nữa ta có đpcm nha

a)Ta có: a³ + b³ + c³ = 3abc

=>a³+b³+c³-3abc=1/2(a+b+c)((a-b)²+(b-c)²+(c-a)²) =0 (dễ dàng phân tích được bạn tự làm)

=>Có 2 trường hợp 

a+b+c=0(loại vì a+b+c khác 0 ) hoặc (a-b)²+(b-c)²+(c-a)² = 0 

Mà (a-b)² , (b-c)² , (c-a)² >= 0 với mọi a,b,c

=>để (a-b)² + (b-c)² + (c-a)² = 0

=>a=b=c

Thay trường hợp a=b=c vào P

=> (2017 +1)(2017+1)(2017+1)=2018³

b)Tương tự a+b+c=0

=> a³ + b³ + c³ = 3abc

=>\(A=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ac}\)

\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)

\(A=\frac{3abc}{abc}=3\) Do (a³ +b³ + c³=3abc thay vào)

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

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

\(\Rightarrow a=b=c=1\)

\(A=\frac{a^{1000}\cdot b^{1007}}{c^{2007}}=\frac{1\cdot1}{1}=1\)

Thay a+b+c=2017 vào \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\)  ta có:

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

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Rightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)\(\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)\(\Rightarrow\left(a+b\right)\left(\frac{c\left(a+b+c\right)+ab}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{c\left(b+c\right)+ca+ab}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+ca+ab\right]=0\)

\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+a\left(b+c\right)\right]=0\)

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

\(\Rightarrow\)\(a+b=0\) hoặc \(b+c=0\) hoặc \(c+a=0\)

\(\Rightarrow\)\(c=2017\)hoặc \(a=2017\) hoặc \(b=2017\left(đpcm\right)\)