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:

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

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

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

1: (a-1)(a-3)(a-4)(a-6)+9

=(a^2-7a+6)(a^2-7a+12)+9

=(a^2-7a)^2+18(a^2-7a)+81

=(a^2-7a+9)^2>=0

b: \(A=\dfrac{a^4-4a^3+a^2+4a^3-16a+4+16a-3}{a^2}=\dfrac{16a-3}{a^2}\)

a^2-4a+1=0

=>a=2+căn 3 hoặc a=2-căn 3

=>A=11-4căn 3 hoặc a=11+4căn 3

Ta có: a+b+c=0

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

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

\(\Leftrightarrow2\left(ab+bc+ac\right)=0-1=-1\)

hay \(ab+bc+ac=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(b+c+a\right)=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)

Ta có: \(M=a^4+b^4+c^4\)

\(\Leftrightarrow M=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2-2a^2b^2-2a^2c^2-2b^2c^2\)

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

\(\Leftrightarrow M=1^2-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy: \(M=\dfrac{1}{2}\)

Ta có : \(a+b+c=0\)

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

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)=1\) ( * )

\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)

Lại có : \(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\) ( suy ra từ * )

\(\Rightarrow a^4+b^4+c^4=2\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

Vậy ...

Bạn ơi! ABC khác 0 thì làm sao ạ+b+c=0 được bạn

           Bài làm :

Vì :

 \(a+b+c=0\)

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

Ta có :

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

\(\Rightarrow A=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}\)

\(\Rightarrow A=\left(-\frac{c}{b}\right).\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)

\(\Rightarrow A=-\frac{abc}{abc}\)

\(\Rightarrow A=-1\)

Vậy A=-1

học sinh giỏi toán đâu hết rồi

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