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

\(a^3-abc+b^3-abc+c^3-abc=0\)

\(a^3+b^3+c^3-3abc=0\)

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

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

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

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

Mà \(a+b+c\ne0\)

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

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

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

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

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

mình làm hơi tắt.

Đến đây bạn tự làm nốt nhé~

Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)

\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)

Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)

\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)

Áp dụng t/c bắc cầu ta dc : \(a=b=c\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)

=>a²+b²=2ab

=>a²-2ab+b²=0

=>(a-b)²=0=>a=b

tương tự=>b=c

=>a=b=c

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)

(a+b+c)(1/a+1/b+1/c)=3a.3/a=9

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

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

\(=3+\frac{a^2+b^2}{ab}+\frac{a^2+c^2}{ac}+\frac{b^2+c^2}{bc}\)

\(=3+\frac{a^2+b^2}{\frac{a^2+b^2}{2}}+\frac{a^2+c^2}{\frac{a^2+c^2}{2}}+\frac{b^2+c^2}{\frac{b^2+c^2}{2}}\)

\(=3+2+2+2=9\)

Có: \(a+b+c=0\)

=> \(a^3+b^3+c^3=3abc\) (tự chứng minh)

\(A=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)

 a² + b² + c² + d² + e² ≥ a(b + c + d + e) 

Ta có: a² + b² + c² + d² + e² 

= (a²/4 + b²) + (a²/4 + c²) + (a²/4 + d²) + (a²/4 + e²) 

Lại có: (a/2 - b)² ≥ 0 <=> a²/4 - ab + b² ≥ 0 <=> a²/4 + b² ≥ ab 

Tương tự ta có: 

. a²/4 + c² ≥ ac 
. a²/4 + d² ≥ ad 
. a²/4 + e² ≥ ae 

--> (a²/4 + b²) + (a²/4 + c²) + (a²/4 + d²) + (a²/4 + e²) ≥ ab + ac + ad + ae 

<=> a² + b² + c² + d² + e² ≥ a(b + c + d + e) --> đ.p.c.m 

Dấu " = " xảy ra <=> a/2 = b = c = d = e 
 

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

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

\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{a^2b}-\frac{3}{ab^2}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)

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

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

\(\Rightarrow\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ac}{b^2}=3\)

ta thấy từ a+b+c=0 \(\Leftrightarrow a^3+b^3+c^3=3abc\)(được cm nhiều trg sách cx như trên mạng)

\(\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)

suy ra đpcm

Ta có : \(a+b+c=0\)

Lập phương 2 vế lên ta có :

\(\left(a+b+c\right)^3=0^3\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

Ta lại có:

\(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}-3=0\)

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

\(\Leftrightarrow\frac{a^3+b^3+c^3}{abc}-3=0\)

Theo chứng minh trên có : \(a^3+b^3+c^3=3abc\)

\(\Rightarrow\frac{3abc}{abc}-3=0\)

\(\Leftrightarrow3-3=0\)( đúng ) 

Vậy với \(a+b+c=0\left(a\ne0;b\ne0;c\ne0\right)\)thì \(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}-3=0\)

Đến chỗ: \(a^3+b^3+c^3=3abc\)

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

=> \(\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3\)

=> \(\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}-3=0\) là đc rồi em nhé!

Dòng thứ 9 trở xuống là khồn đúng đâu nhé!

Cho mình sửa đề một chút nha

\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)(*)

Theo bài ra , ta có :

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

\(\Leftrightarrow a+b=-c\)

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

\(\Leftrightarrow a^2+2ab+b^2=c^2\)

\(\Leftrightarrow a^2+b^2-c^2=-2ab\) (1)

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

\(\Leftrightarrow b+c=-a\)

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

\(\Leftrightarrow b^2+2bc+c^2=a^2\)

\(\Leftrightarrow b^2+c^2-a^2=-2bc\) (2)

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

\(\Leftrightarrow a+c=-b\)

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

\(\Leftrightarrow a^2+2ac+c^2=b^2\)

\(\Leftrightarrow a^2+c^2-b^2=-2ac\) (3)

Thay (1) , (2) , (3) vào (*) ta được

\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)

\(=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}+-\dfrac{1}{2}+-\dfrac{1}{2}=-\dfrac{3}{2}\)

Vậy \(A=-\dfrac{3}{2}\)

Chúc bạn học tốt =))

A=\(\dfrac{-7}{5}\)

\(a+b=-c\Rightarrow a^2+b^2+2ab=c^2\Rightarrow a^2+b^2-c^2=-2ab\)
\(\frac{ab}{a^2+b^2-c^2}=\frac{ab}{-2ab}=-\frac{1}{2}\)
tương tự \(\Rightarrow A=\frac{-3}{2}\)