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Ta có:\(a^2-b=b^2-c\)
\(\Leftrightarrow a^2-b^2=b-c\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)
\(\Leftrightarrow a+b=\frac{b-c}{a-b}\)
\(\Leftrightarrow a+b+1=\frac{b-c}{a-b}+1\)
\(\Leftrightarrow a+b+1=\frac{a-c}{a-b}\)
Cmtt ta có:
\(\hept{\begin{cases}b^2-c=c^2-a\Leftrightarrow b+c+1=\frac{b-a}{b-c}\\c^2-a=a^2-b\Leftrightarrow c+a+1=\frac{c-b}{c-a}\end{cases}}\)
\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-c}{b-a}.\frac{c-b}{c-a}=-1\)
Cre:mạng
Đặt x=b-c; y=c-a; z=a-b
=>x+y+z=b-c+c-a+a-b=0
Ta có: \(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b} = 0\)
=>\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
=>ayz+bxz+cxy=0
=>\(a=\frac{(y-z)x^2}{y^2 + yz + z^2};\quad b=\frac{(z-x)y^2}{z^2 + zx + x^2};\quad c=\frac{(x-y)z^2}{x^2 + xy + y^2}\)
A=\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{a}{x^2}+\frac{b}{y^2}+\frac{c}{z^2}\)
\(=\frac{1}{x^2}\cdot\frac{(y-z)x^2}{y^2 + yz + z^2}+\frac{1}{y^2}\cdot\frac{(z-x)y^2}{z^2 + zx + x^2}+\frac{1}{z^2}\cdot\frac{(x-y)z^2}{x^2 + xy + y^2}\)
\(=\frac{y-z}{y^2 + yz + z^2}+\frac{z-x}{z^2 + zx + x^2}+\frac{x-y}{x^2 + xy + y^2}\)
x+y+z=0
=>x+y=-z; x+z=-y; y+z=-x
\(y^2 + yz + z^2 = (y + z)^2 - yz = (-x)^2 - yz = x^2 - yz\)
\(z^2 + zx + x^2 = y^2 - zx\)
\(x^2 + xy + y^2 = z^2 - xy\)
\(A=\frac{y-z}{x^2 - yz}+\frac{z-x}{y^2 - zx}+\frac{x-y}{z^2 - xy}\)
\(=\frac{(y-z)(y^2 - zx)(z^2 - xy) + (z-x)(z^2 - xy)(x^2 - yz) + (x-y)(x^2 - yz)(y^2 - zx)}{(x^2 - yz)(y^2 - zx)(z^2 - xy)}\)
=0
=>ĐPCM
Vì a + b + c = 0
<=> (a + b + c)2 = 0
<=> a2 + b2 + c2 = -2(ab + bc + ca)
Khi đó \(\frac{9\left(a^2+b^2+c^2\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}=\frac{-18\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(=\frac{-18\left(ab+bc+ca\right)}{-6\left(ab+bc+ca\right)}=3\)