http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/

Ta có: \(a+b+c=0\)

\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)

\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)

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

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

\(=\frac{3}{2}\)

A=3

B=3

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)

\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)

\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)

     Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)

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

\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)

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

Sau đó bạn thực hiện tiếp nhé.

Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)

Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)

Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)

Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)

Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)

theo bài ra ta có:

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

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

=> \(\frac{abc}{ca+cb}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)

vì a,b,c khác 0 => ca+cb = ab+ac = bc+ba

=> a = b = c

ta có:

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

vậy M = 1

Mik ko bk đúng hay sai đâu nha!

Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath

Học tốt=)

tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2

Ukm,mình không để ý.Sorry bn

Ta có:\(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)

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

Tươmg tự ta cũng có:\(b^2+c^2-a^2=-2bc\) và \(c^2+a^2-b^2=-2ca\)

\(\Rightarrow P=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ca}=-\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0\)

a+b+c=0 =>  a= -(b+c) TƯƠNG TỰ

                    b= -(a+c) ; c= -(b+a)

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

 =>   P= \(\frac{1}{2c\left(b+c\right)}+\frac{1}{2b\left(a+c\right)}+\frac{1}{2a\left(b+c\right)}​\)

 thay b+c=-a; a+c=-b ; a+b=-c (như trên )

=> P= \(\frac{1}{-2ac}+\frac{1}{-2ab}+\frac{1}{-2bc}\)

 QUY ĐONG CAC MAU THUC TA CO 

P= \(\frac{a+b+c}{-2abc}\)

a+b+c=0 => P=0