Mình sửa lại đề nha:    \(A=2x^2+2y^2+2xy-2x+2y+2\)

Ta có:  \(A=2x^2+2y^2+2xy-2x+2y+2\)

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

                \(=\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\)\(\ge0\)

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

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

\(\Leftrightarrow2A=4x^2+4y^2-4xy-4x+4y+4\)

\(=\left(4x^2-4xy+y^2\right)-2\left(2x-y\right)+1+3y^2+6y+4\)

\(=\left(\left(2x-y\right)^2-2\left(2x-y\right)+1\right)+\left(3y^2+6y+3\right)+2\)

\(=\left(2x-y-1\right)^2+3\left(y+1\right)^2+2\ge2\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}y+1=0\\2x-y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}y=-1\\x=0\end{cases}}\)

4M = 4x^2+4y^2-4xy+8x-16y-8072

= [(4x^2-4xy+y^2)-2.(2x+y).2+4]+(3y^2-12y+12)-8088

= [(2x-y)^2-2.(2x-y).2+4]+3.(y^2-4y+4)-8088

= (2x-y-2)^2+3.(y-2)^2-8088 >= -8088

=> M >= -2022

Dấu "=" xảy ra <=> 2x-y-2=0 và y-2=0 <=> x=y=2

Vậy GTNN của M = -2022 <=> x=y=2

Tk mk nha 

các phương trình bậc nhất là a) c) d) 

Các phương trình bậc nhất:

a: 1 + x = 0

c: 1 - 2t = 0

d: 3y = 0

a,\(x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)2x-2x+2\(x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)

=\(\left(x^4-x^3+2x^2-2x+2\right)\left(x-1\right)\)

b,

câu b, c,d tương tự 

bạn tự làm nó nhé

\(pt\Leftrightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}.\)

\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)

Mà \(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}#0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy nghiệm của pt là x = -1

dkxd  \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)

b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)

\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)

tới khúc này bí rồi ^^

a,ĐKXĐ của A là:\(x\ne+2;-2\)

b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)

c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)

Ta có bảng

Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))