\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
\(=\sqrt{\dfrac{\left(bc\right)^2+\left(ac\right)^2+\left(ab\right)^2}{\left(abc\right)^2}}\)
\(=\dfrac{\sqrt{\left(bc+ac+ab\right)^2-2abc\left(a+b+c\right)}}{abc}\)
(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