Giúp em với ạ: Tìm các số nguyên dương x, y, z thỏa mãn xz=y^2 và x^2+z^2+99=7y^2. 26/08/2021 Bởi Ximena Giúp em với ạ: Tìm các số nguyên dương x, y, z thỏa mãn xz=y^2 và x^2+z^2+99=7y^2.
Đáp án: $\begin{array}{l} + )xz = {y^2} \Rightarrow 2xz = 2{y^2}\\ + ){x^2} + {z^2} + 99 = 7{y^2}\\ \Rightarrow {x^2} + {z^2} + 2xz + 99 = 7{y^2} + 2{y^2}\\ \Rightarrow {\left( {x + z} \right)^2} + 99 = 9{y^2} = {\left( {3y} \right)^2}\\ \Rightarrow {\left( {x + z} \right)^2} – {\left( {3y} \right)^2} = – 99\\ \Rightarrow \left( {x + z + 3y} \right)\left( {x + z – 3y} \right) = – 99 = – \left( {9.11} \right) = – \left( {3.33} \right) = – \left( {99.1} \right)\\Goi\,x + z = a;\,3y = b\\ \Rightarrow \left( {a + b} \right)\left( {a – b} \right) = – \left( {9.11} \right) = – \left( {3.33} \right) = – \left( {99.1} \right)\\ + TH1:\left( {a + b} \right)\left( {a – b} \right) = – \left( {9.11} \right)\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a + b = 11\\a – b = – 9\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 10\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + z = 1\\3y = 10\end{array} \right.\left( {ktm\,} \right)\\\left\{ \begin{array}{l}a + b = 9\\a – b = – 11\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = – 1\\b = 10\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + z = – 1\\3y = 10\end{array} \right.\left( {ktm\,} \right)\end{array} \right.\\ + TH2:\left( {a + b} \right)\left( {a – b} \right) = – \left( {3.33} \right)\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a + b = 33\\a – b = – 3\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 15\\b = 18\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + z = 15\\3y = 18\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + z = 15\\y = 6 \Rightarrow xz = {6^2} = 36\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 12\\z = 3\\y = 6\end{array} \right.\left( {tm} \right)\\\left\{ \begin{array}{l}a + b = 3\\a – b = – 33\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = – 15\\b = 18\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x + z = – 15\\3y = 18\end{array} \right.\left( {ktm\,} \right)\end{array} \right.\\ + TH3:không\,thỏa\,mãn\\Vay\,x = 12;y = 6;z = 3\,Hoặc\,x = 3;y = 6;z = 12\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
+ )xz = {y^2} \Rightarrow 2xz = 2{y^2}\\
+ ){x^2} + {z^2} + 99 = 7{y^2}\\
\Rightarrow {x^2} + {z^2} + 2xz + 99 = 7{y^2} + 2{y^2}\\
\Rightarrow {\left( {x + z} \right)^2} + 99 = 9{y^2} = {\left( {3y} \right)^2}\\
\Rightarrow {\left( {x + z} \right)^2} – {\left( {3y} \right)^2} = – 99\\
\Rightarrow \left( {x + z + 3y} \right)\left( {x + z – 3y} \right) = – 99 = – \left( {9.11} \right) = – \left( {3.33} \right) = – \left( {99.1} \right)\\
Goi\,x + z = a;\,3y = b\\
\Rightarrow \left( {a + b} \right)\left( {a – b} \right) = – \left( {9.11} \right) = – \left( {3.33} \right) = – \left( {99.1} \right)\\
+ TH1:\left( {a + b} \right)\left( {a – b} \right) = – \left( {9.11} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a + b = 11\\
a – b = – 9
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b = 10
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 1\\
3y = 10
\end{array} \right.\left( {ktm\,} \right)\\
\left\{ \begin{array}{l}
a + b = 9\\
a – b = – 11
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = – 1\\
b = 10
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = – 1\\
3y = 10
\end{array} \right.\left( {ktm\,} \right)
\end{array} \right.\\
+ TH2:\left( {a + b} \right)\left( {a – b} \right) = – \left( {3.33} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a + b = 33\\
a – b = – 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 15\\
b = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 15\\
3y = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 15\\
y = 6 \Rightarrow xz = {6^2} = 36
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 12\\
z = 3\\
y = 6
\end{array} \right.\left( {tm} \right)\\
\left\{ \begin{array}{l}
a + b = 3\\
a – b = – 33
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = – 15\\
b = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = – 15\\
3y = 18
\end{array} \right.\left( {ktm\,} \right)
\end{array} \right.\\
+ TH3:không\,thỏa\,mãn\\
Vay\,x = 12;y = 6;z = 3\,Hoặc\,x = 3;y = 6;z = 12
\end{array}$