Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\Rightarrow ab+bc+ac=0$

Đặt $ab=x, bc=y, ac=z$ thì $x+y+z=0$

Có:

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

$=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y)^3-3xy(x+y)+z^3}{xyz}$

$=\frac{(-z)^3-3xy(-z)+z^3}{xyz}$
$+\frac{-z^3+3xyz+z^3}{xyz}=\frac{3xyz}{xyz}=3$

a+b+c=0

=>a+b=-c; a+c=-b; b+c=-a

\(a^2-b^2-c^2\)

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

\(=a^2-\left\lbrack\left(b+c\right)^2-2bc\right\rbrack=a^2-\left(a^2-2bc\right)=2bc\)

\(b^2-a^2-c^2\)

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

\(=b^2-\left\lbrack\left(a+c\right)^2-2ac\right\rbrack=b^2-\left(b^2-2ac\right)=2ac\)

\(c^2-a^2-b^2\)

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

\(=c^2-\left\lbrack\left(a+b\right)^2-2ab\right\rbrack=c^2-\left(c^2-2ab\right)=2ab\)

Ta có: \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)

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

\(=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2bac}=\frac{\left(-c\right)^3-3ab\cdot\left(-c\right)+c^3}{2abc}=\frac{3abc}{2abc}=\frac32\)

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

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

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

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

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

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

Ta có: \(M=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

ĐK : a;b;c khác 0 

Thấy : \(a^2+b^2+c^2=\left(a+b+c\right)^2\Leftrightarrow ab+bc+ac=0\) (1)

Ta có : \(P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)

Từ (1) suy ra : \(\left(b+c\right)a=-bc\Leftrightarrow\dfrac{b+c}{a}=\dfrac{-bc}{a^2}\)   

CMTT ; ta có : \(\dfrac{c+a}{b}=\dfrac{-ac}{b^2};\dfrac{a+b}{c}=\dfrac{-ab}{c^2}\)

Suy ra : \(P=-\left(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\right)=-\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^2b^2c^2}\)  (2) 

Đặt : ab = x ; bc = y ; ac = z ; ta có : x + y + z = 0 \(\Rightarrow x^3+y^3+z^3=3xyz\)  (3)

Từ (2) và (3) suy ra : \(P=-\dfrac{3xyz}{xyz}=-3\)

Vậy ... 

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Lời giải:

\(A=\frac{(bc)^3+(2ac)^3+(2ab)^3}{8a^2b^2c^2}=\frac{(bc)^3+(2ac+2ab)^3-3.2ac.2ab(2ac+2bc)}{8a^2b^2c^2}\)

\(=\frac{(bc)^3+(-bc)^3+12a^2b^2c^2}{8a^2b^2c^2}=\frac{12}{8}=1,5\)

từ giả thiết ta có

a+b+c=0

<=>  a=-(b+c0

         a²=b²  +c² +2bc

tương tự   b²=a²+c²+2ac

                c²=a²+b²+2ab

thay vào Q ta đc

\(Q=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)

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

\(Q=\frac{-1}{2ab}-\frac{1}{2bc}-\frac{1}{2ac}\)

\(Q=\frac{-b-a-c}{2abc}\)

\(Q=\frac{-\left(a+b+c\right)}{2abc}\)

\(Q=0\)

Vậy với a,b,c khác 0, a+b+c=0 thì Q=0

a+b+c=0

=>a+b=-c; a+c=-b; b+c=-a

\(a^2+b^2-c^2\)

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

\(b^2+c^2-a^2\)

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

\(=\left(-a\right)^2-2bc-a^2=-2bc\)

\(c^2+a^2-b^2\)

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

\(=\left(-b\right)^2-2ac-b^2=-2ac\)

Ta có: \(H=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)

\(=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}=-\frac12-\frac12-\frac12=-\frac32\)

\(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(a+c\right)^2-2ac-b^2}=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(a+b+c\right)\left(b+c-a\right)-2bc}+\dfrac{ac}{\left(a+b+c\right)\left(a+c-b\right)-2ac}=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}.3=-\dfrac{3}{2}\)