Bài 1: Chắc đề là \(a^4-4a^2+4a-1\)
Ta có: \(a^4-4a^2+4a-1=a^4-\left(4a^2-4a+1\right)=a^4-\left(2a-1\right)^2=\left(a^2-2a+1\right)\left(a^2+2a-1\right)=\left(a-1\right)^2\left(a^2+2a-1\right)\)
là số nguyên tố \(\Leftrightarrow\left[{}\begin{matrix}\left(a-1\right)^2=1\\a^2+2a-1=1\end{matrix}\right.\) ( tự giải tiếp)
Bài 2: Làm mẫu một bài thôi nhé
a) Đặt A = \(2x^2+2y^2+2xy-8x-10y+2025\)
\(2A=4x^2+4y^2+4xy-16x-20y+4050\)
\(=\left(2x\right)^2+2.2x\left(y-4\right)+\left(y-4\right)^2-\left(y-4\right)^2+4y^2-20y+4050\)
\(=\left(2x+y-4\right)^2-\left(y^2-8y+16\right)+4y^2-20y+4050\)
\(=\left(2x+y-4\right)^2+3y^2-12y+4034=\left(2x+y-4\right)^2+3\left(y^2-4y+4\right)+4022=\left(2x+y-4\right)^2+3\left(y-2\right)^2+4022\ge4022\forall x,y\)
Vậy min A = 4022 \(\Leftrightarrow\left\{{}\begin{matrix}2x+y-4=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(B=\frac{1}{2}\left(x^2+4y^2+64+4xy-16x-32y\right)+\frac{3}{2}x^2-6x+1988\)
\(B=\frac{1}{2}\left(x+2y-8\right)^2+\frac{3}{2}\left(x-2\right)^2+1992\ge1982\)
\(B_{min}=1982\) khi \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
