theo giả thiết => a-c=\(2\sqrt{2}\)
ta có: \(2\cdot A=2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
=\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
=\(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+8\)
=14
=> A=7
\(\left\{{}\begin{matrix}a-b=\sqrt{2}+1\\b-c=\sqrt{2}-1\end{matrix}\right.\Rightarrow a-c=2\sqrt{2}\)
\(A=a^2+b^2+c^2-ab-bc-ca\)
\(2A=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2\)
\(=2+2\sqrt{2}+1+2-2\sqrt{2}+1+8\)
\(=14\)
Vậy \(A=7\)
\(a-b=\sqrt{2}+1;b-c=\sqrt{2}-1\)
=>\(a-b+b-c=\sqrt{2}+1+\sqrt{2}-1\)
\(\Leftrightarrow a-c=2\sqrt{2}\)
Ta có:
\(A^2=2a^2+2b^2+2c^2-2ab-2bc-2ca\)
\(A^2=a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2\)
\(A^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
\(A^2=\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2\)
\(A^2=3+2\sqrt{2}+3-2\sqrt{2}+8=14\)
A=\(\sqrt{14}\)
làm lại
từ đầu bài suy ra \(a-b+b-c=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
2A=\(2a^2+2b^2+2c^2-2ab-2bc-2ca\)
\(2A=a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\)
\(2A=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
\(2A=\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2\)
2A =\(3+2\sqrt{2}+3-2\sqrt{2}+8=14\)
suy ra A=7
