\(A=\left(2018-2016\right)\left(2018+2016\right)=2.4034\)

\(B=\left(2019-2017\right)\left(2019+2017\right)=2.4036\)

Ta thấy 4034 < 4036 nên A < B.

\(A=2018^2-2016^2=\left(2018+2016\right)\left(2018-2016\right)=4034.2\)

\(B=2019^2-2017^2=\left(2019+2017\right)\left(2019-2017\right)=4036.2\)

Vì 4036 > 4034 nên 4036 . 2 > 4034 . 2 nên B > A

2017.2019 = (2018-1)(2018+1) = 2018² -1 => a =1

b= 2018³ +1 (???)

b) \(-4x^2-4x-1\)

\(=-\left(4x^2+4x+1\right)\)

\(=-\left(2x+1\right)^2\)

c) \(\frac{4}{9}x^2-25y^2\)

\(=\left(\frac{2}{3}x+5y\right)\left(\frac{2}{3}x-5y\right)\)

d) \(\frac{1}{27}x^3-8\)

\(=\left(\frac{1}{3}x-2\right)\left(\frac{1}{9}x+\frac{2}{3}x+4\right)\)

Đề sai không bạn

ko sai nhưng khó

\(A=x^2+2y^2+2xy+2x-4y+2016\)

\(=x^2+y^2+y^2+2xy+2x+2y-6y+2016\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+\left(2x+2y\right)+2007\)

\(=\left(x+y\right)^2+\left(y-3\right)^2+2\left(x+y\right)+2007\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)

Vì \(\hept{\begin{cases}\left(x+y+1\right)^2\ge0;\forall x,y\\\left(y-3\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2+2006\ge0+2006;\forall x,y\)

Hay \(A\ge2006;\forall x,y\)

Dấu"=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy \(A_{min}=2006\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Mình làm có gì sai hả @@ 

biết làm luôn rồi :)

Ta có hệ phương trình : 

\(\hept{\begin{cases}x^2+7=4y^2+4y\left(1\right)\\x^2+3xy+2y^2+x+y=0\left(2\right)\end{cases}}\)

Từ (2) \(\Leftrightarrow x^2+xy+2xy+2y^2+x+y=0\)

\(\Leftrightarrow\left(x+y\right)\left(x+2y+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-y\\x=-2y-1\end{cases}}\)

*) Với \(x=-y\) thì từ (1) suy ra :

\(\left(-y\right)^2+7=4y^2+4y\)

\(\Leftrightarrow3y^2+4y-7=0\)

\(\Leftrightarrow\left(y-1\right)\left(3y+7\right)=0\Leftrightarrow\orbr{\begin{cases}y=1\\y=-\frac{7}{3}\end{cases}}\)

+) Khi \(y=1\Rightarrow x=-1\)

+) Khi \(y=-\frac{7}{3}\Rightarrow x=\frac{7}{3}\)

*) Với \(x=-2y-1\) thì từ (1) suy ra :

\(\left(-2y-1\right)^2+7=4y^2+4y\)

\(\Leftrightarrow4y^2+4y+1+7=4y^2+4y\)

\(\Leftrightarrow0=8\) ( Vô lí )

Vậy \(\left(x,y\right)\in\left\{\left(-1,1\right);\left(\frac{7}{3},-\frac{7}{3}\right)\right\}\)

Với mọi x, y

A chia hết cho B

<=> \(x^4y^3+3x^3y^3+x^2y^n⋮4x^ny^2\)

Khi đó: \(x^4;x^3;x^2⋮x^n\Rightarrow n\le2\)

\(y^3;y^n⋮y^2\Rightarrow n\ge2\)

Từ 2 điều trên => n = 2.

nhanh