Question

[Sao Vậy Trời]

cho a,b,c là số hữu tỉ ,a=b+c

cm a) \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)  là số hữu tỉ

      b) \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\)   là số hữu tỉ

Answer 1

[alibaba nguyễn]

a/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)

\(=\sqrt{\frac{\left(b+c\right)^2.b^2+\left(b+c\right)^2.c^2+b^2.c^2}{\left(b+c\right)^2.b^2.c^2}}\)

\(=\sqrt{\frac{\left(b^2+bc+c^2\right)^2}{\left(b+c\right)^2.b^2.c^2}}\)

\(=\left|\dfrac{b^2+bc+c^2}{\left(b+c\right).b.c}\right|\)

Vậy \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)là số hữu tỉ

Answer 2

[alibaba nguyễn]

b/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}+\frac{1}{\left(2b+c\right)^2}}\)

\(=\sqrt{\frac{\left(b+c\right)^2.b^2+\left(b+c\right)^2.\left(2b+c\right)^2+\left(2b+c\right)^2.b^2}{\left(b+c\right)^2.\left(2b+c\right)^2.b^2}}\)

\(=\sqrt{\frac{\left(3b^2+3bc+c^2\right)^2}{\left(b+c\right)^2.\left(2b+c\right)^2.b^2}}\)

\(=\left|\dfrac{3b^2+3bc+c^2}{\left(b+c\right).\left(2b+c\right).b}\right|\)

Vậy \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\) là số hữu tỉ