1 Tìm x ; y ; z biết :
a, Cho \(\dfrac{x-2}{4}\) = \(\dfrac{y+1}{5}\)= \(\dfrac{z+3}{7}\) và 2x+ y - 2 z = 7
b, Cho \(\dfrac{x}{3}\) = \(\dfrac{y-2}{5}\) = \(\dfrac{z+3}{7}\) và 2x - y - z = 17
c, Cho \(\dfrac{x}{y}\) = \(\dfrac{5}{4}\) ; \(\dfrac{y}{z}\) = \(\dfrac{2}{3}\) và \(x^2\) - \(2y^2\) - \(3z^2\) = - 460
e, Cho 2x = 3y ; 4y = 5z và xyz = 9600
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
