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Có:
\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)
Thay \(a=b=c\) vào \(A\), ta được:
\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)
\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)
\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)
\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)
\(=\dfrac{3}{2017^2}\)
Vậy: ...
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 \(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)
a: Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a+b+c=0\)
a) \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)(đúng do a+b+c = 0)
a: Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow a+b+c=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-\left(a+b+c\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{ac+bc+c^2}\)
\(\Leftrightarrow-\left(a+b\right)ab=\left(a+b\right)\left(ac+bc+c^2\right)\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+\left(a+b\right)ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
=> 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
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
\(\Rightarrow a=b=c=1\)
\(A=\frac{a^{1000}\cdot b^{1007}}{c^{2007}}=\frac{1\cdot1}{1}=1\)
bài 1
ab+bc+ca=0
=>ab+bc=-ca
ta có (a+b)(b+c)(c+a)/abc
=> (ab+ac+bc+b2)(c+a)/abc
=> (0+b2)(c+a)/abc
=>b2c+b2a/abc
=>b(ab+bc)/abc
=>b(-ac)/abc
=>-abc/abc=-1
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)