Đáp án:$\left( {x;y} \right) \in \left\{ {\left( {2; - 1} \right);\left( { - 2; - 5} \right);\left( { - 4; - 3} \right)} \right\}$

Giải thích các bước giải:

$\begin{array}{l}
a)P.3x{y^2} = 3{x^2}{y^3} + 6{x^2}{y^2} + 3x{y^3} + 6x{y^2}\\
 \Rightarrow P.3x{y^2} = 3x{y^2}.xy + 3x{y^2}.2x + 3x{y^2}.y + 3x{y^2}.2\\
 \Rightarrow P.3x{y^2} = 3x{y^2}\left( {xy + 2x + y + 2} \right)\\
 \Rightarrow P = \frac{{3x{y^2}\left( {xy + 2x + y + 2} \right)}}{{3x{y^2}}}\left( {dk:x \ne 0;y \ne 0} \right)\\
 \Rightarrow P = xy + 2x + y + 2\\
b)P = 3\\
 \Rightarrow xy + 2x + y + 2 = 3\\
 \Rightarrow x\left( {y + 2} \right) + y + 2 = 3\\
 \Rightarrow \left( {y + 2} \right)\left( {x + 1} \right) = 3 = 1.3 = \left( { - 1} \right).\left( { - 3} \right)\left( {do:x,y \in Z} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 1\\
y + 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 3\\
y + 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 =  - 1\\
y + 2 =  - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 =  - 3\\
y + 2 =  - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0;y = 1\\
x = 2;y =  - 1\\
x =  - 2;y =  - 5\\
x =  - 4;y =  - 3
\end{array} \right.\\
Mà:x \ne 0;y \ne 0\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2; - 1} \right);\left( { - 2; - 5} \right);\left( { - 4; - 3} \right)} \right\}
\end{array}$

Đáp án:(x;y)∈{(2;−1);(−2;−5);(−4;−3)}(x;y)∈{(2;−1);(−2;−5);(−4;−3)}

Giải thích các bước giải:

a)P.3xy2=3x2y3+6x2y2+3xy3+6xy2⇒P.3xy2=3xy2.xy+3xy2.2x+3xy2.y+3xy2.2⇒P.3xy2=3xy2(xy+2x+y+2)⇒P=3xy2(xy+2x+y+2)3xy2(dk:x≠0;y≠0)⇒P=xy+2x+y+2b)P=3⇒xy+2x+y+2=3⇒x(y+2)+y+2=3⇒(y+2)(x+1)=3=1.3=(−1).(−3)(do:x,y∈Z)⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{x+1=1y+2=3{x+1=3y+2=1{x+1=−1y+2=−3{x+1=−3y+2=−1⇒⎡⎢ ⎢ ⎢ ⎢⎣x=0;y=1x=2;y=−1x=−2;y=−5x=−4;y=−3Mà:x≠0;y≠0⇒(x;y)∈{(2;−1);(−2;−5);(−4;−3)}a)P.3xy2=3x2y3+6x2y2+3xy3+6xy2⇒P.3xy2=3xy2.xy+3xy2.2x+3xy2.y+3xy2.2⇒P.3xy2=3xy2(xy+2x+y+2)⇒P=3xy2(xy+2x+y+2)3xy2(dk:x≠0;y≠0)⇒P=xy+2x+y+2b)P=3⇒xy+2x+y+2=3⇒x(y+2)+y+2=3⇒(y+2)(x+1)=3=1.3=(−1).(−3)(do:x,y∈Z)⇒[{x+1=1y+2=3{x+1=3y+2=1{x+1=−1y+2=−3{x+1=−3y+2=−1⇒[x=0;y=1x=2;y=−1x=−2;y=−5x=−4;y=−3Mà:x≠0;y≠0⇒(x;y)∈{(2;−1);(−2;−5);(−4;−3)

  5 sao nhé