abc=a+b+c => 1 = 1/ab + 1/bc + 1/ac 

3 = 1/a+1/b+1/c => 5 = 1/a^2 + 1/b^2 + 1/c^2 + 2/ab + 2/ac + 2/cb 

=> 5 = 1/a^2 + 1/b^2 + 1/c^2 + 2(1/ab + 1/ac + 1/bc) = M + 2 

=> M = 5 - 2 = 3

thank bạn

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

\(\Leftrightarrow ab+bc+ca=0\)

Ta có: \(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)

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

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

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

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

PS: Hồi tối lười để người khác làm mà không ai làm thôi t làm vậy

( a+b+c)^2 = a^2 + b^2 + c^2 

=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = a^2 + b^2 + c^2 

=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ac - a^2 - b^2 - c^2 = 0 

=> 2ab + 2bc + 2ac = 0 

ta có 

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

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

đến đây bạn nhóm lại nhé mk giải ra thì dài lắm nên chỉ gợi ý cho bn đấy đây thôi

Các cậu giúp mình với.Sắp nộp bài rổi

đơn giản, cứ áp dụng theo công thức là ra!!!!

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

=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)

=>\(\dfrac{ab+ac+bc}{abc}=0\)

=>ab+ac+bc=0

=>ab=-ac-bc

ac=-ab-bc

bc=-ab-ac

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

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

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

N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)

N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0