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\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
\(P=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) hinh nhu theo co dieu kien a,b,c ko dong thoi = 0
<=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
<=> \(\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)
<=> \(\left(a+b\right)\left(ac+bc+c^2\right)=-ab\left(a+b\right)\)
<=> \(\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)
<=> \(\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
<=> \(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
<=> a+b=0 hoac a+c=0 hoac b+c=0
do khi luy thua a,b,c len cach so mu le la 27,41,2019 thi a,b,c ko doi dau nen \(a^{27}+b^{27}=0.hoac.b^{41}+c^{41}=0.hoac.c^{2019}+a^{2019}=0\)
P = 0
Vay P = 0
Study well
Ta có : \(\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{a}\Rightarrow\frac{b+c}{bc}=\frac{a-a-b-c}{a^2+ab+ac}\)
\(\Leftrightarrow\frac{b+c}{bc}=\frac{-b-c}{a^2+ab+ac}\Leftrightarrow\left(b+c\right)\left(a^2+ab+ac\right)=-\left(b+c\right)bc\)
\(\left(b+c\right)\left(a^2+ab+ac\right)+\left(b+c\right)bc=0\)
\(\Rightarrow\left(b+c\right)\left(a^2+ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(b+c\right)[\left(a+b\right)a+c\left(a+b\right)]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\c^{2019}+a^{2019}=0\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\a^{2019}+c^{2019}=0\end{cases}}\end{cases}}}\)
Ta có : \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Do a;b;c đôi một khác nhau nên \(\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ne0\)
Do đó a + b + c = 0
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ta có :
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)
\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự : \(\hept{\begin{cases}M.\frac{a}{b-c}=1+\frac{2a^3}{abc}\\M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\end{cases}}\)
Cộng vế với vế ta được \(P=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a=b=c\Rightarrow P=2020^3\)
TH2: \(a+b+c=0\) ko đủ dữ kiện tính ra giá trị cụ thể của P