Question

Tìm 2 số x,y biết x/2=y/5 ; xy - 15x + 6y = 40

Answer 1

Đặt x/2=y/5=k

=>x=2k; y=5k

xy-15x+6y=40

\(\Leftrightarrow10k^2-15\cdot2k+6\cdot5k=40\)

\(\Leftrightarrow10k^2=40\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

=>x=4;y=10

TRường hợp 2: k=-2

=>x=-4; y=-10

Answer 2

Đặt `x/2 = y/5 = k`

`=>` `{(x = 2k),(y = 5k):}`

Ta có `: xy - 15x + 6y = 40`

`=> 2k . 5k - 15 . ( 2k ) + 6 . ( 5k ) = 40`

`=> 10k^2 - 30k + 30k = 40`

`=> k^2 = 40 : 10`

`=> k^2 = 4`

`=>` \(\left[ \begin{array}{l}k^2 = 2^2\\k^2 = ( - 2 )^2\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}k = 2\\k = - 2\end{array} \right.\) 

Xét `k = 2 => {(x = 2 . 2 = 4),(y = 5 . 2 = 10):}`

Xét `k = - 2 => {(x = - 2 . 2 = - 4),(y = - 2 . 5= - 10):}`

Vậy `, ( x ; y ) in { ( 4 ; 10 ) ; ( - 4 ; - 10 ) } .`

 

Answer 3

` x/2=y/5 => x=2y /5`

thay` x = 2y /5 `vào `xy - 15x + 6y = 40` ta đc

\(\dfrac{2}{5}y^2-6y+6y=40\)

\(=>\dfrac{2}{5}y^2=40\)

\(=>y^2=100\Leftrightarrow\left[{}\begin{matrix}y=10\\y=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{5}.10=4\\x=\dfrac{2}{5}.\left(-10\right)=-4\end{matrix}\right.\)