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

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

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

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

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

=> a=-b hoặc b=-c hoặc c=-a

không mất tính tổng quát ,giả sử a=-b, ta có:

\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{-b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\left(1\right)\)

\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\left(2\right)\)

Từ  (1) và (2) => đpcm

Tương tự với 2 trường hợp còn lại ta cũng có đpcm

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\)\(\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\)\(\left(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\right)+\left(\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}\right)=0\)

\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

Ta có : 

\(\frac{1}{a^2+b^2+c^2}< \frac{1}{a^2};\frac{1}{b^2};\frac{1}{c^2}\)

\(\Rightarrow\)\(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}< 0\)

\(\Rightarrow\)\(x^2=y^2=z^2=0\)\(\Rightarrow\)\(x=y=z=0\) ( đpcm ) 

Chúc bạn học tốt ~ 

a. ĐK: a, b, c khác 0.

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

\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)

\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)

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

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).

b) Không mất tính tổng quả. G/s: a = b + c

Khi đó ta có:

\(\frac{a^2+b^2-c^2}{2ab}=\frac{\left(b+c\right)^2+b^2-c^2}{2\left(b+c\right)b}=1\)

\(\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b+c\right)^2}{2bc}=-1\)

\(\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+\left(b+c\right)^2-b^2}{2\left(b+c\right)c}=1\)

=> Điều phải chứng minh.

Ta có : \(y+z=ax+cz+ax+by=2ax+x\)

\(\Rightarrow\)\(y+z-x=2ax\)\(\Rightarrow\)\(a=\frac{y+z-x}{2x}\)\(\Rightarrow\)\(\frac{1}{a+1}=\frac{2x}{x+y+z}\)

Tương tự, ta cũng có \(\frac{1}{b+1}=\frac{2y}{x+y+z};\frac{1}{c+1}=\frac{2z}{x+y+z}\)

\(\Rightarrow\)\(S=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Chúc bạn học tốt ~ 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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