Bài tập: P.3xy^2=3x^2y^3+6x^2y^2+3xy^3+6xy^2 a) Tìm P b) Tìm cặp số nguyên xy để P=3 08/08/2021 Bởi Hailey Bài tập: P.3xy^2=3x^2y^3+6x^2y^2+3xy^3+6xy^2 a) Tìm P b) Tìm cặp số nguyên xy để P=3
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 Bình luận
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: $\begin{array}{l}a)P.3x{y^2} = 3{x^2}{y^3} + 6{x^2}{y^2} + 3x{y^3} + 6x{y^2}\\ \Rightarrow P.3x{y^2} = 3x{y^2}.xy + 3x{y^2}.2x + 3x{y^2}.y + 3x{y^2}.2\\ \Rightarrow P.3x{y^2} = 3x{y^2}\left( {xy + 2x + y + 2} \right)\\ \Rightarrow P = \frac{{3x{y^2}\left( {xy + 2x + y + 2} \right)}}{{3x{y^2}}}\left( {dk:x \ne 0;y \ne 0} \right)\\ \Rightarrow P = xy + 2x + y + 2\\b)P = 3\\ \Rightarrow xy + 2x + y + 2 = 3\\ \Rightarrow x\left( {y + 2} \right) + y + 2 = 3\\ \Rightarrow \left( {y + 2} \right)\left( {x + 1} \right) = 3 = 1.3 = \left( { – 1} \right).\left( { – 3} \right)\left( {do:x,y \in Z} \right)\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 1 = 1\\y + 2 = 3\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = 3\\y + 2 = 1\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = – 1\\y + 2 = – 3\end{array} \right.\\\left\{ \begin{array}{l}x + 1 = – 3\\y + 2 = – 1\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = 0;y = 1\\x = 2;y = – 1\\x = – 2;y = – 5\\x = – 4;y = – 3\end{array} \right.\\Mà:x \ne 0;y \ne 0\\ \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2; – 1} \right);\left( { – 2; – 5} \right);\left( { – 4; – 3} \right)} \right\}\end{array}$ Bình luận
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
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:
$\begin{array}{l}
a)P.3x{y^2} = 3{x^2}{y^3} + 6{x^2}{y^2} + 3x{y^3} + 6x{y^2}\\
\Rightarrow P.3x{y^2} = 3x{y^2}.xy + 3x{y^2}.2x + 3x{y^2}.y + 3x{y^2}.2\\
\Rightarrow P.3x{y^2} = 3x{y^2}\left( {xy + 2x + y + 2} \right)\\
\Rightarrow P = \frac{{3x{y^2}\left( {xy + 2x + y + 2} \right)}}{{3x{y^2}}}\left( {dk:x \ne 0;y \ne 0} \right)\\
\Rightarrow P = xy + 2x + y + 2\\
b)P = 3\\
\Rightarrow xy + 2x + y + 2 = 3\\
\Rightarrow x\left( {y + 2} \right) + y + 2 = 3\\
\Rightarrow \left( {y + 2} \right)\left( {x + 1} \right) = 3 = 1.3 = \left( { – 1} \right).\left( { – 3} \right)\left( {do:x,y \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 1\\
y + 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 3\\
y + 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = – 1\\
y + 2 = – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = – 3\\
y + 2 = – 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0;y = 1\\
x = 2;y = – 1\\
x = – 2;y = – 5\\
x = – 4;y = – 3
\end{array} \right.\\
Mà:x \ne 0;y \ne 0\\
\Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2; – 1} \right);\left( { – 2; – 5} \right);\left( { – 4; – 3} \right)} \right\}
\end{array}$