\(a^2+b^2+c^2+d^2=1\) và \(ab+bc+cd+da=1\)
\(\Rightarrow a^2+b^2+c^2+d^2=ab+bc+cd+da\)
\(\Rightarrow a^2+b^2+c^2+d^2-ab-bc-cd-da=0\)
\(\Rightarrow2\left(a^2+b^2+c^2+d^2-ab-bc-cd-da\right)=0.2\)
\(\Rightarrow2a^2+2b^2+2c^2+2d^2-2ab-2bc-2cd-2da=0\)
\(\Rightarrow a^2+a^2+b^2+b^2+c^2+c^2+d^2+d^2-2ab-2bc-2cd-2da=0\)
\(\Rightarrow\left(a^2-2ab-b^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2cd+d^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(c-d\right)^2=0\)
Ta có:
\(\left(a-b\right)^2\ge0\)
\(\left(a-d\right)^2\ge0\)
\(\left(b-c\right)^2\ge0\)
\(\left(c-d\right)^2\ge0\)
Mà tổng của chúng đều là 0
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow a-d=0\Rightarrow a=d\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow c-d=0\Rightarrow c=d\)
\(\Rightarrow a=b=c=d\)
Thay: \(a^2+b^2+c^2+d^2=1\) ta được
\(\Rightarrow a^2+a^2+a^2+a^2=1\)
\(\Rightarrow4a^2=1\)
\(\Rightarrow a^2=\frac{1}{4}\)
\(\Rightarrow a\in\left\{\pm\frac{1}{2}\right\}\)
