Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

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

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\(xy+3x+4y=x\left(y+3\right)+4y=5\Leftrightarrow x\left(y+3\right)+4y+12=17\Leftrightarrow\left(x+4\right)\left(y+3\right)=17\)

\(2xy+x-2y-1=3\Leftrightarrow x\left(2y+1\right)-\left(2y+1\right)=3\Leftrightarrow\left(x-1\right)\left(2y+1\right)=3\)

g. G(x)=2x²+2y²+z²+2xy-2xz-2yz-2x-4y

           = [x²+2x(y-z)+(y²-2yz+z²)]+(x²-2x+1)+(y²-4y+4)-5

          = (x+y-z)²+(x-1)²+(y-2)²-5

Vì (x+y-z)²≥0∀x,y,z

     (x-1)²≥0∀x

      (y-2)²≥0∀y

⇒ G  = (x+y-z)²+(x-1)²+(y-2)²-5 ≥ -5

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=3\\x=1\\y=2\end{matrix}\right.\)

h,H(x)=x² + y²-xy-x+y+1

⇔ 2H=2x²+2y²-2xy-2x-2y+2

         = (x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)

         = (x-y)²+(x-1)²+(y-1)²

Vì (x-y)²≥0 ∀x,y

    (x-1)²≥0 ∀x

     (y-1)² ≥0 ∀y

⇒ 2H≥0 ⇒ H≥0

Dấu "=" xảy ra ⇔ x=y=1

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)

Ta có bảng:

Vậy không có x,y thỏa mãn đề bài 

b, tương tự câu a

 \(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)

Rồi làm tương tự câu a

\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)

Rồi làm tương tự câu a