Ta có: x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4

xy=xy24⇔6x24=xy24xy=xy24⇔6x24=xy24

⇒6x=xy⇒6x=xy

⇒y=6⇒y=6

x−61=x+67x−61=x+67

⇔7.(x−6)=x+6⇔7.(x−6)=x+6

⇔7x−42=x+6⇔7x−42=x+6

⇔7x−x=6+42⇔7x−x=6+42

⇔6x=48⇔6x=48

⇒x=8⇒x=8

Vậy x=8;y=6

xy=48

48

xy=48

48

\(x^2+y^2+1=xy-x-y\Leftrightarrow2x^2+2y^2+2=2xy-2x-2y\)

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

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

\(A=\frac{1}{xy}+2\left(x+y\right)=\frac{1}{\left(-1\right)\left(-1\right)}+2\left[\left(-1\right)+\left(-1\right)\right]=\frac{-7}{2}\)

Dễ ợt ak

Hướng dẫn thôi nhé:

Lời giải:

a)\(xy+x+y+1=0\)

\(\Rightarrow x\left(y+1\right)+1\left(y+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=0\)

b)\(xy-x-y=0\)

\(\Rightarrow xy-x-y+1=1\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

c)\(xy-x-y-1=0\)

\(\Rightarrow xy-x-y+1=2\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=2\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=2\)

d) \(xy-x-y+1=0\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)

e)\(xy+2x+y+11=0\)

\(\Rightarrow xy+2x+y+2=-9\)

\(\Rightarrow x\left(y+2\right)+1\left(y+2\right)=-9\)

\(\Rightarrow\left(x+1\right)\left(y+2\right)=-9\)

Lời giải:
a.

$P+Q=2x^4+xy(x^2+y^2)-2ax(x+y)+1+x^4+xy^3+3x^2-xy+3$

$=2x^4+x^3y+xy^3-2ax^2-2axy+1+x^4+xy^3+3x^2-xy+3$

$=3x^4+x^3y+2xy^3-(2a-3)x^2-(2a+1)xy+4$

b. 

$P-Q=2x^4+xy(x^2+y^2)-2ax(x+y)+1-(x^4+xy^3+3x^2-xy+3)$

$=2x^4+x^3y+xy^3-2ax^2-2axy+1-x^4-xy^3-3x^2+xy-3$

$=x^4+x^3y-(2a+3)x^2-(2a-1)xy-2$