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a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> ab + bc + ca = \(\frac{a^2+b^2+c^2}{2}\)
=> \(\left(ab+bc+ca\right)^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)
=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2a^2bc+2ab^2c+2abc^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)
=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)
=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\left(\frac{a^2+b^2+c^2}{2}\right)^2\)(vì a + b + c = 0)
Lại có \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{a^2b^2+b^2c^2+a^2c^2}{a^2b^2c^2}=\frac{\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2}{\left(abc\right)^2}\)
\(=\frac{\left(\frac{a^2+b^2+c^2}{2}\right)^2}{\left(abc\right)^2}=\left(\frac{\frac{a^2+b^2+c^2}{2}}{abc}\right)^2=\left(\frac{a^2+b^2+c^2}{2abc}\right)^2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)là bình phương của 1 số hữu tỉ
3/ Ta có:
\(x+y+z=0\)
\(\Rightarrow x^2=\left(y+z\right)^2;y^2=\left(z+x\right)^2;z^2=\left(x+y\right)^2\)
\(a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
Ta có:
\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)
\(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\)
\(=-ax^2-by^2-cz^2\)
\(\Leftrightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Leftrightarrow ax^2+by^2+cz^2=0\)
1/ Đặt \(a-b=x,b-c=y,c-z=z\)
\(\Rightarrow x+y+z=0\)
Ta có:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{a+b+c}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{0}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
