bạn học đến đg tròn rồi à

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)  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 : \(\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{a}\Rightarrow\frac{b+c}{bc}=\frac{a-a-b-c}{a^2+ab+ac}\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\c^{2019}+a^{2019}=0\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\a^{2019}+c^{2019}=0\end{cases}}\end{cases}}}\)

\(VT=\sqrt{-\left(x-1\right)^2+3}+\sqrt{-\left(x+3\right)^2+1}< 1+\sqrt{3}\) pt vô nghiệm 

Sang h.vn mà hỏi đây là trang hỏi t,v,a mà (chủ yếu là toán)