\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) 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 : \(\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{a}\Rightarrow\frac{b+c}{bc}=\frac{a-a-b-c}{a^2+ab+ac}\)
\(\Leftrightarrow\frac{b+c}{bc}=\frac{-b-c}{a^2+ab+ac}\Leftrightarrow\left(b+c\right)\left(a^2+ab+ac\right)=-\left(b+c\right)bc\)
\(\left(b+c\right)\left(a^2+ab+ac\right)+\left(b+c\right)bc=0\)
\(\Rightarrow\left(b+c\right)\left(a^2+ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(b+c\right)[\left(a+b\right)a+c\left(a+b\right)]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\c^{2019}+a^{2019}=0\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\a^{2019}+c^{2019}=0\end{cases}}\end{cases}}}\)
\(\Rightarrow\)b= -c hoặc a= -b hoặc c= -a
\(\Rightarrow\)\(b^{41}+c^{41}=0\)hoặc \(a^{27}+b^{27}=0\)hoặc \(c^{2019}+a^{2019}=0\)
Trong 3 trường hợp trên , thì P luôn bằng 0 do có một thừa số bằng 0
Vậy P =0
