Toán tìm x,y,z :2x=3y,2y=5z và x^2 -2y^2+yz=260 10/10/2021 By Aubrey tìm x,y,z :2x=3y,2y=5z và x^2 -2y^2+yz=260
Đáp án: $\begin{array}{l}2x = 3y \Rightarrow \dfrac{x}{3} = \dfrac{y}{2} \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}}\\2y = 5z \Rightarrow \dfrac{y}{5} = \dfrac{z}{2} \Rightarrow \dfrac{y}{{10}} = \dfrac{z}{4}\\ \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{4} = k\\ \Rightarrow \left\{ \begin{array}{l}x = 15.k\\y = 10.k\\z = 4.k\end{array} \right.\\Khi:{x^2} – 2{y^2} + y.z = 260\\ \Rightarrow {\left( {15k} \right)^2} – 2.{\left( {10k} \right)^2} + 10.k.4.k = 260\\ \Rightarrow 225.{k^2} – 200.{k^2} + 40{k^2} = 260\\ \Rightarrow 65{k^2} = 260\\ \Rightarrow {k^2} = 4\\ \Rightarrow \left[ \begin{array}{l}k = 2\\k = – 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 30;y = 20;z = 8\\x = – 30;y = – 20;z = – 8\end{array} \right.\\Vậy\,\left( {x;y;z} \right) = \left\{ {\left( {30;20;8} \right);\left( { – 30; – 20; – 8} \right)} \right\}\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
2x = 3y \Rightarrow \dfrac{x}{3} = \dfrac{y}{2} \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}}\\
2y = 5z \Rightarrow \dfrac{y}{5} = \dfrac{z}{2} \Rightarrow \dfrac{y}{{10}} = \dfrac{z}{4}\\
\Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{4} = k\\
\Rightarrow \left\{ \begin{array}{l}
x = 15.k\\
y = 10.k\\
z = 4.k
\end{array} \right.\\
Khi:{x^2} – 2{y^2} + y.z = 260\\
\Rightarrow {\left( {15k} \right)^2} – 2.{\left( {10k} \right)^2} + 10.k.4.k = 260\\
\Rightarrow 225.{k^2} – 200.{k^2} + 40{k^2} = 260\\
\Rightarrow 65{k^2} = 260\\
\Rightarrow {k^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
k = 2\\
k = – 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 30;y = 20;z = 8\\
x = – 30;y = – 20;z = – 8
\end{array} \right.\\
Vậy\,\left( {x;y;z} \right) = \left\{ {\left( {30;20;8} \right);\left( { – 30; – 20; – 8} \right)} \right\}
\end{array}$