a. Theo đề bài ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\) và 5x-y+3z=124
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=31\Rightarrow x=31.3=93\\\dfrac{y}{5}=31\Rightarrow y=31.5=155\\\dfrac{z}{-2}=31\Rightarrow z=\left(-2\right).31=-62\end{matrix}\right.\)
Vậy.........
\(a,x:y:z=5:3:\left(-2\right)\)và \(5x-y+3z=124\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{\left(-2\right)}\Rightarrow\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{\left(-6\right)}\)
\(=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)
\(\Rightarrow x=31.3=93\)
\(y=31.5=155\)
\(z=31.\left(-2\right)=\left(-62\right)\)
Vậy........
\(b,x:y=3:7\)và \(x^2-2y^2=\left(-356\right)\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x^2}{9}=\dfrac{2y^2}{98}\)
\(\dfrac{x^2-2y^2}{9-98}=\dfrac{\left(-356\right)}{\left(-89\right)}=4\)
\(\Rightarrow x=4.3=12\)
\(y=4.7=28\)
Vậy.......
