Vì a,b,c,d có vai trò như nhau

Giả sử \(a\ge b\ge c\ge d\)

=>\(a^2\ge b^2\ge c^2\ge d^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\le\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}\)

=>\(1\le4.\frac{1}{d^2}\)

=>\(\frac{1}{4}\le\frac{1}{d^2}\)

=>\(4\ge d^2\)

=>\(2\ge d\)

Vì d là số tự nhiên khác 0

=>d=1,2

-Xét d=1

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{1^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=0\)

Vì\(\frac{1}{a^2}>0,\frac{1}{b^2}>0,\frac{1}{c^2}>0=>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}>0\)

=>Vô lí

-Xét d=2

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{2^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{4}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)

Vì \(a\ge b\ge c\)

=>\(a^2\ge b^2\ge c^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}\)

=>\(\frac{3}{4}\le3.\frac{1}{c^2}\)

=>\(\frac{1}{4}\le\frac{1}{c^2}\)

=>\(4\ge c^2\)

=>\(2\ge c\)

Vì \(c\ge d=>c\ge2\)

=>c=2

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{2^2}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{4}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)

Vì \(a\ge b\)

=>\(a^2\ge b^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}\le\frac{1}{b^2}+\frac{1}{b^2}\)

=>\(\frac{2}{4}\le\frac{2}{b^2}\)

=>\(\frac{1}{4}\le\frac{1}{b^2}\)

=>\(4\ge b^2\)

=>\(2\ge b\)

Vì \(b\ge c=>b\ge2\)

=>b=2

=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{2^2}=\frac{2}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{4}=\frac{2}{4}\)

=>\(\frac{1}{a^2}=\frac{1}{4}\)

=>\(a^2=4=>a=2\)

Vậy a=2,b=2,c=2,d=2