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1.
Cho $a+b+c=0$.
Ta có:
$a+b+c=0\Rightarrow a=-(b+c)$
$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$
$\Rightarrow a^2-b^2-c^2=2bc$
Tương tự:
$b^2-c^2-a^2=2ca$
$c^2-a^2-b^2=2ab$
Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$
$=\dfrac{a^3+b^3+c^3}{2abc}$
Mà $a+b+c=0$ nên:
$a^3+b^3+c^3=3abc$
Suy ra $A=\dfrac{3abc}{2abc}$
$A=\dfrac32$
2.
Cho $abc=2$.
Ta có:
$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$
Vì $abc=2$ nên:
$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$
Suy ra:
$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$
$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$
Quy đồng và sử dụng $abc=2$:
$A=1$
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)
Thay (*) vào M ta được:
\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)
\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)
Vậy M = 0
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ac+ba}{abc}=0\Leftrightarrow bc+ac+ba=0\Leftrightarrow c.\left(a+b\right)=-ba\Leftrightarrow a+b=\frac{-ab}{c}\)
\(b+c=-\frac{bc}{a},a+c=\frac{-ac}{b}\)
thay vô là đc :") lazzy~~