a.
Đặt \(n^2+391=k^2\) với \(k\in Z\)
\(\Rightarrow k^2-n^2=391\)
\(\Rightarrow\left(k-n\right)\left(k+n\right)=391\)
k-n | 391 | -391 | 1 | -1 | 17 | -17 | 23 | -23 |
k+n | 1 | -1 | 391 | -391 | 23 | -23 | 17 | -17 |
n | -195 | 195 | 195 | -195 | 3 | -3 | -3 | 3 |
Vậy \(n=\left\{\pm195;\pm3\right\}\)
b.
Với mọi số thực a; b ta luôn có:
\(2\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2\ge0\)
\(\Leftrightarrow2a^2-4ab+2b^2+a^2-4a+4+b^2-4b+4\ge0\)
\(\Leftrightarrow3\left(a^2+b^2\right)+8\ge4\left(a+b+ab\right)\)
\(\Leftrightarrow3\left(a^2+b^2\right)+8\ge32\)
\(\Leftrightarrow3\left(a^2+b^2\right)\ge24\)
\(\Rightarrow a^2+b^2\ge8\) (đpcm)
Dấu "=" xảy ra khi \(a=b=2\)