Question

Answer 1

Có ai biết làm ko

Answer 2

Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\Rightarrow x=4k;y=7k\)

Ta có \(xy=252\Rightarrow28k^2=252\Rightarrow k^2=9\)

\(\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12;y=21\\x=-12;y=-21\end{matrix}\right.\)

Answer 3

\(\dfrac{x}{4}=\dfrac{y}{7}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)

\(\Rightarrow xy=28k^2=252\Rightarrow k=\pm3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm21\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=21\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-21\end{matrix}\right.\end{matrix}\right.\)

Answer 4

Nhân cả 2 vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{7}=\dfrac{xy}{4.7}=\dfrac{252}{28}=9\)

\(\left(\dfrac{x}{4}\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=3\\\dfrac{x}{4}=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)

Thay x vào biểu thức \(\dfrac{x}{4}=\dfrac{y}{7}\), ta có: \(\Rightarrow\left[{}\begin{matrix}\dfrac{12}{4}=\dfrac{y}{7}\\\dfrac{-12}{4}=\dfrac{y}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=21\\y=-21\end{matrix}\right.\)

 

 

Answer 5