Bài 1:
\(\dfrac{-3x^4y^5+8x^{2n}y^6}{4x^{n+1}\cdot y^n}=-\dfrac{3}{4}\cdot x^{4-n-1}\cdot y^{5-n}+\dfrac{8}{4}\cdot x^{2n-n-1}\cdot y^{6-n}\)
\(=-\dfrac{3}{4}x^{3-n}y^{5-n}+2x^{n-1}y^{6-n}\)
Để đây là phép chia hết thì \(\left\{{}\begin{matrix}3-n>=0\\5-n>=0\\n-1>=0\\6-n>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}n< =3\\n< =5\\n>=1\\n< =6\end{matrix}\right.\Leftrightarrow1< =n< =3\)
mà n là số tự nhiên
nên \(n\in\left\{1;2;3\right\}\)
Bài 2:
a: \(A=16x^2+5y^2-8x+2023\)
\(=16x^2-8x+1+5y^2+2022\)
\(=\left(4x-1\right)^2+5y^2+2022>=2022\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}4x-1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=0\end{matrix}\right.\)
b: \(B=\left(x-3\right)^2+\left(x-1\right)^2\)
\(=x^2-6x+9+x^2-2x+1\)
\(=2x^2-8x+10\)
\(=2\left(x^2-4x+5\right)=2\left(x^2-4x+4+1\right)\)
\(=2\left(x-2\right)^2+2>=2\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2