## tìm x,y,z :2x=3y,2y=5z và x^2 -2y^2+yz=260

Question

tìm x,y,z :2x=3y,2y=5z và x^2 -2y^2+yz=260

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2 tháng 2021-10-10T15:08:08+00:00 1 Answers 6 views 0

## Answers ( )

1. Đáp án:

$\begin{array}{l} 2x = 3y \Rightarrow \dfrac{x}{3} = \dfrac{y}{2} \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}}\\ 2y = 5z \Rightarrow \dfrac{y}{5} = \dfrac{z}{2} \Rightarrow \dfrac{y}{{10}} = \dfrac{z}{4}\\ \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{4} = k\\ \Rightarrow \left\{ \begin{array}{l} x = 15.k\\ y = 10.k\\ z = 4.k \end{array} \right.\\ Khi:{x^2} – 2{y^2} + y.z = 260\\ \Rightarrow {\left( {15k} \right)^2} – 2.{\left( {10k} \right)^2} + 10.k.4.k = 260\\ \Rightarrow 225.{k^2} – 200.{k^2} + 40{k^2} = 260\\ \Rightarrow 65{k^2} = 260\\ \Rightarrow {k^2} = 4\\ \Rightarrow \left[ \begin{array}{l} k = 2\\ k = – 2 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = 30;y = 20;z = 8\\ x = – 30;y = – 20;z = – 8 \end{array} \right.\\ Vậy\,\left( {x;y;z} \right) = \left\{ {\left( {30;20;8} \right);\left( { – 30; – 20; – 8} \right)} \right\} \end{array}$