help me!!!
Ta có:
\(\frac{x}{2}=\frac{y}{5}\) => \(\frac{x^2}{4}=\frac{y^2}{25}=\frac{x.y}{2.5}=\frac{10}{10}=1\)
=> \(\begin{cases}x^2=1.4.=4\\y^2=1.25=25\end{cases}\)=> \(\begin{cases}x\in\left\{2;-2\right\}\\y\in\left\{5;-5\right\}\end{cases}\)
Vậy \(\begin{cases}x=2\\y=5\end{cases}\); \(\begin{cases}x=-2\\y=-5\end{cases}\)
Ta có:\(\left[\begin{array}{nghiempt}\frac{x}{2}=\frac{y}{5}\\x.y=10\end{array}\right.\)
Đặt: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k;y=5k\)
\(\Rightarrow2k.5k=10\)
\(\Rightarrow10k^2=10\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=1\)
\(\Rightarrow x=2k\Rightarrow x=2.1=2\)
\(\Rightarrow y=5k\Rightarrow y=5.1=5\)
Đặt \(\frac{x}{2}=\frac{y}{5}=k\)
Ta có : \(\frac{x}{2}=2k;\frac{y}{5}=5k\)
\(\Rightarrow2k.5k=10\)
\(\Rightarrow10k^2=10\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
Nếu \(k=1\) , thì :
\(\begin{cases}\frac{x}{2}=1\Rightarrow x=2\\\frac{y}{5}=1\Rightarrow y=5\end{cases}\)
Nếu \(k=-1\) , thì :
\(\begin{cases}\frac{x}{2}=-1\Rightarrow x=-2\\\frac{y}{5}=-1\Rightarrow y=-5\end{cases}\)
Ta có: \(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{2}=\frac{xy}{5}=\frac{10}{5}=2\)
=> x2 = 2.2 = 4
=> x \(\in\left\{2;-2\right\}\)
=> Nếu x = 2 thì y = 10 : 2 = 5
Nếu x = - 2 thì y = 10 : (- 2) = - 5
Vậy x \(\in\left\{2;-2\right\};y\in\left\{5;-5\right\}\)
