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\(\left\{{}\begin{matrix}x-y=a\\y-z=b\\z-x=c\end{matrix}\right.\Leftrightarrow a+b+c=0\)
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y-z\right)^2}+\dfrac{1}{\left(z-x\right)^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}=\dfrac{\left(ab+bc+ac\right)^2-2abc\left(a+b+c\right)}{a^2b^2c^2}\)
\(=\left(\dfrac{ab+bc+ac}{abc}\right)^2=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) là bp 1 số hữu tỉ(đpcm)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
\(=\sqrt{\dfrac{\left(bc\right)^2+\left(ac\right)^2+\left(ab\right)^2}{\left(abc\right)^2}}\)
\(=\dfrac{\sqrt{\left(bc+ac+ab\right)^2-2abc\left(a+b+c\right)}}{abc}\)
(áp dụng HĐT: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)\))
\(=\dfrac{\sqrt{\left[a\left(b+c\right)+bc\right]^2-2abc\left[a+\left(b+c\right)\right]}}{abc}\)
\(=\dfrac{\sqrt{\left(a^2+bc\right)^2-4a^2bc}}{abc}\)
\(=\dfrac{\sqrt{a^4+2a^2bc+\left(bc\right)^2-4a^2bc}}{abc}\)
\(=\dfrac{\sqrt{a^4-2a^2bc+\left(bc\right)^2}}{abc}\)
\(=\dfrac{a^2-bc}{abc}\) là 1 số hữu tỉ (đpcm)
Ta có:
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(=\dfrac{\left(b+c\right)^2b^2+\left(b+c\right)^2c^2+b^2c^2}{b^2c^2\left(b+c\right)^2}\)
\(=\dfrac{b^4+2b^3c+3b^2c^2+2bc^3+c^4}{b^2c^2\left(b+c\right)^2}\)
\(=\dfrac{\left(b^4+2b^2c^2+c^4\right)+2bc\left(b^2+c^2\right)+b^2c^2}{b^2c^2\left(b+c\right)^2}\)
\(=\dfrac{\left(b^2+bc+c^2\right)^2}{b^2c^2\left(b+c\right)^2}\)
\(\Rightarrow\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\dfrac{\left(b^2+bc+c^2\right)^2}{b^2c^2\left(b+c\right)^2}}=\dfrac{b^2+bc+c^2}{bc\left(b+c\right)}\)
Vì a, b, c là các số hữu tỉ nên \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\) là số hữu tỉ
Đặt \(\left\{{}\begin{matrix}\dfrac{a}{b^2}=x\\\dfrac{b}{c^2}=y\\\dfrac{c}{a^2}=z\end{matrix}\right.\Rightarrow xyz=1;x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
Ta có \(x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Leftrightarrow x+y+z=xy+yz+zx\)
\(\Leftrightarrow xyz-1+x+y+z-xy-yz-zx=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b^2}=1\\\dfrac{b}{c^2}=1\\\dfrac{c}{a^2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b^2\\b=c^2\\c=a^2\end{matrix}\right.\left(đpcm\right)\)
\(A=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}=2+\dfrac{1}{4}=\dfrac{9}{4}\)