a) Ta có: \(\frac{x}{5}=\frac{y}{4}\)
\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{16}\)
Ta có: \(x^2-y^2=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)
Do đó:
\(\left\{{}\begin{matrix}\frac{x^2}{25}=\frac{1}{9}\\\frac{y^2}{16}=\frac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=\frac{25}{9}\\y^2=\frac{16}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{\frac{5}{3};-\frac{5}{3}\right\}\\y\in\left\{\frac{4}{3};-\frac{4}{3}\right\}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{3};-\frac{5}{3}\right\}\) và \(y\in\left\{\frac{4}{3};-\frac{4}{3}\right\}\)
b) Ta có: \(\frac{x}{y}=\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}\)
Ta có: \(x^2+y^2=208\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{208}{13}=16\)
Do đó:
\(\left\{{}\begin{matrix}\frac{x^2}{4}=16\\\frac{y^2}{9}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=64\\y^2=144\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{8;-8\right\}\\y\in\left\{12;-12\right\}\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-8\right\}\) và \(y\in\left\{12;-12\right\}\)
a/ Đặt: \(\frac{x}{5}=\frac{y}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
Ta có: \(x^2-y^2=1\)
\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)
\(\Rightarrow25k^2-16k^2=1\)
\(\Rightarrow k^2\left(25-16\right)=1\)
\(\Rightarrow k^29=1\)
\(\Rightarrow k^2=\frac{1}{9}\)
\(\Rightarrow k=\pm\sqrt{\frac{1}{9}}=\pm\frac{1}{3}\)
*Với: \(k=\frac{1}{3}\)
\(\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.\frac{1}{3}=\frac{5}{3}\\y=4.\frac{1}{3}=\frac{4}{3}\end{matrix}\right.\)
*Với: \(k=-\frac{1}{3}\)
\(\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.\left(-\frac{1}{3}\right)=-\frac{5}{3}\\y=4\left(-\frac{1}{3}\right)=-\frac{4}{3}\end{matrix}\right.\)
Vậy:..................
b/ \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Đặt: \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Ta có: \(x^2+y^2=208\)
\(\Rightarrow\left(2k\right)^2+\left(3k\right)^2=208\)
\(\Rightarrow4k^2+9k^2=208\)
\(\Rightarrow k^2\left(4+9\right)=208\)
\(\Rightarrow k^213=208\)
\(\Rightarrow k^2=208:13=16\)
\(\Rightarrow k=\pm4\)
*Với k = 4
\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=3.4=12\end{matrix}\right.\)
*Với k = -4
\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2.\left(-4\right)=-8\\y=3.\left(-4\right)=-12\end{matrix}\right.\)
Vậy...................
