\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Rightarrow2\left(ab+bc+ac\right)=0\)

\(\Rightarrow ab+bc+ac=0\)

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

\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)

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

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

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)

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

\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Ta có: a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)

<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

<=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Mà a+b+c khác 0

=>\(a^2+b^2+c^2-ab-bc-ca=0\)

<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)

=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

\(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\)

\(\Rightarrow\frac{1}{x\left(x^2+1\right)}=\frac{a\left(x^2+1\right)}{x\left(x^2+1\right)}+\frac{x\left(bx+c\right)}{x\left(x^2+1\right)}\)

\(\Rightarrow a\left(x^2+1\right)+x\left(bx+c\right)=1\)

\(\Rightarrow ax^2+a+xbx+xc=1\)

sai đề rồi cậu ơi! I là trung điểm của AC rồi đằng sau I còn là trung điểm của HC, CE

post lại đề,khó jiểu quá 

cho m,n là các số nguyên dương thỏa mãn

 mn+1 chia hết cho 24 cm m+n chia hết cho 24

\(2x\cdot\left(x^2-7x-3\right)=2x^4-14x^2-6x\)