\(\dfrac{x}{2}=\dfrac{y}{3}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}\)
\(=\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{54}{6}=9\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{4}=9\\\dfrac{y^2}{9}=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=6\\y=9\end{matrix}\right.;\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
\(\dfrac{x}{5}=\dfrac{y}{3}\)
\(\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2}{25}=\dfrac{y^2}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=\dfrac{1}{4}\\\dfrac{y^2}{9}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4x^2=25\\4y^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.;\left\{{}\begin{matrix}x=\dfrac{-5}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
a) Ta có :
\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)và xy= 54
=> \(\dfrac{x.x}{2}\)=\(\dfrac{x.y}{3}\)và x.y=54
=>\(\dfrac{x^2}{2}\)=\(\dfrac{54}{3}\)
=> \(^{x^2}\)=18
=> \(^{x^2}\)= 18.2
=> \(^{x^2}\)=36
=> x = 6 và x = -6
=> y = 6 va y =-9
b)
Ta có
\(\dfrac{x}{5}\)= \(\dfrac{y}{3}\) và\(^{x^2}\)-\(^{y^2}\)= 4
=> \(\dfrac{x^2}{25}\)= \(\dfrac{y^2}{9}\) và \(^{x^2}\)- \(^{y^2}\)=4
Theo tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x^2}{25}\)= \(\dfrac{y^2}{9}\)= \(\dfrac{x^2-^{y^2}}{25-9}\)=\(\dfrac{4}{16}\)=4
=> \(\dfrac{x}{25}\)=4 => x = 100
=>\(\dfrac{y}{9}\)=4 => y = 36
vậy x = 100 ; y =36
