Question

Answer 1

a, \(\left\{{}\begin{matrix}x+y=-12\\x=\dfrac{20}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{y}+y=-12\\x=\dfrac{20}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20+y^2=-12y\\x=\dfrac{20}{y}\end{matrix}\right.\)

=> y = -10 ; y = -2 

Với y = -10 => x = -2 

Với y = -2 => x = -10

b, \(\left\{{}\begin{matrix}x-y=5\\x=\dfrac{14}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{14}{y}-y=5\\x=\dfrac{14}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14-y^2=5y\\x=\dfrac{14}{y}\end{matrix}\right.\)

=> y = 2 ; y = -7 

Với y = 2 => x = 7 

Với y = -7 => x = -2 

Answer 2

Hello bn 

Answer 3

a. \(\left\{{}\begin{matrix}x+y=-12\\xy=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy+y^2=-12y\\xy=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=-12y-20\\x+y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2+12y+20=0\\x+y=-12\end{matrix}\right.\)

Bn giải tiếp PT bậc 2 ở trên là tìm ra đc nhé

Answer 4

a. <=> x = \(\dfrac{20}{y}\)

pt tương đương :

<=> \(\dfrac{20}{y}+y=-12\)

<=>\(y^2+12y+20=0\)\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-2\end{matrix}\right.\)

b. <=> x=\(\dfrac{14}{y}\)

pt tương đương:

<=> \(\dfrac{14}{y}-y=5\)

<=> \(-y^2-5y+14=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

Answer 5

\(a,\left\{{}\begin{matrix}x+y=-12\\x.y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(-12-y\right)\\\left(-12-y\right).y=20\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-12-y\\-y^2-12y-20=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-12-y\\-y^2-10y-\left(2y+20\right)=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-12-y\\-y.\left(y+10\right)-2.\left(y+10\right)=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-12-y\\\left(y+10\right).\left(-y-2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-12-y\\\left[{}\begin{matrix}y=-10\\y=-2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-10\\x=-2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-2\\x=-10\end{matrix}\right.\end{matrix}\right.\)