a) Ta có :
\(5z^2-3x^2-2y^2=594\)
\(x:y:z=3:4:5\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Leftrightarrow\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{5z^2}{125}\)
Aps dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{5z^2}{125}=\dfrac{5z^2-3x^2-2y^2}{125-27-32}=\dfrac{594}{66}=9\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x^2}{27}=9\Leftrightarrow x^2=81\Leftrightarrow x\in\left\{9;-9\right\}\\\dfrac{2y^2}{32}=9\Leftrightarrow y^2=288\Leftrightarrow y\in\left\{12;-12\right\}\\\dfrac{5z^2}{125}=9\Leftrightarrow z^2=225\Leftrightarrow z\in\left\{15;-15\right\}\end{matrix}\right.\)
Vậy ..............
. Nốt câu b theo đề cậu đã sửa nhé.
\(x+y=x:y=3.\left(x-y\right)\)
\(\Rightarrow x+y=3x-3y\)
\(\Rightarrow y+3y=3x-x\)
\(\Rightarrow4y=2x\)
\(\Rightarrow2y=x\)
\(\Rightarrow x:y=2\)
\(\Rightarrow x+y=2y+y=2\)
\(\Rightarrow3y=2\)
\(\Rightarrow y=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy \(x=\dfrac{4}{3}\) , \(y=\dfrac{2}{3}\) .
nhầm ý b,
b, x = y = x : y = 3(x - y )
