Toán Lớp 7: Tìm x, y, z 3x = 5y = 10z và x^2 – 2y^2 + z^2 =148

TRẢ LỜI

1. Cách 1:

   $\begin{cases} 3x=5y\\3x=10z \end{cases}$ <=> $\begin{cases} y=\dfrac{3}{5}x\\z=\dfrac{3}{10} x\end{cases}$

   =>x^2-2*9/25x^2+9/100x^2=148

   <=>37/100x^2=148<=>x^2=400

   => $\left[\begin{matrix} x=20\\ x=-20\end{matrix}\right.$ => $\left[\begin{matrix} y=12;z=6\\ y=-12;z=-6\end{matrix}\right.$

   <=> Vậy (x,y,z)=(+-20,+-12+-6)

   Cách 2:

   y=3/5x;z=3/10x

   =>x^2-2y^2+z^2=148

   =>x^2-2*9/25x^2+9/100x^2=148

   =>(1-18/25+9/100)x^2=148

   =>37/100 => $\left[\begin{matrix} x^2=148\\ x^2=400 \\x=\pm20\end{matrix}\right.$

   => $\left[\begin{matrix} y=\pm12\\ z=\pm6\end{matrix}\right.$

   Trả lời
2. $ 3x = 5y = 10z ⇒ 3x/30 = 5y/30 = 10z/30 $ 

   $ ⇒ x/10 = y/6 = z/3 = k $

   $ ⇒ $ \(\left[ \begin{array}{l}x=10k\\y=6k\\z=3k\end{array} \right.\) 

   $ x^2 – 2y^2 + z^2 = 148 $ 

   $ ⇒ ( 10k )^2 – 2 ( 6k )^2 + ( 3k )^2 = 148 $

   $ 10^2 . k^2 – 2 . 6^2 . k^2  + 3^2 . k ^2 = 148 $

   $ 100 . k^2 – 2 . 36 . k^2  + 9 . k^2 = 148 $

   $ 100 . k^2 – 72 . k^2 + 9 . k^2 = 148 $

   $ ( 100 – 72 + 9 ) . k^2 = 148 $

   $ 37 . k^2  = 148 $

   $ k^2 = 148 : 37 = 4 $ 

   $ ⇒ $ \(\left[ \begin{array}{l}k=2\\k=-2\end{array} \right.\) 

   $ TH1 : k = 2 $

   $ ⇒ $ \(\left[ \begin{array}{l}x=10k=10.2=20\\y=6k=6.2=12\\z=3k=3.2=6\end{array} \right.\) 

   $ TH2 : k = -2 $ 

   $ ⇒ $ \(\left[ \begin{array}{l}x=10k=10.(-2)=-20\\y=6k=6.(-2)=-12\\z=3k=3.(-2)=-6\end{array} \right.\) 

   $ ( x,y,z ) = ( 20;12;6 ) ; ( -20;-12;-6 ) $

   Trả lời