Question

Tìm x :

\((\dfrac{7}{2}x+2).(\dfrac{-5}{2}x+5)=0\)

Answer 1

\(\left(\dfrac{7}{2}x+2\right).\left(\dfrac{-5}{2}x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{2}x+2=0\\\dfrac{-5}{2}x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{2}x=-2\\\dfrac{-5}{2}x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-4}{7}\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-4}{7};2\right\}.\)

Answer 2

(7/2x+2).(-5/2x+5)=0

*7/2x+2=0

7/2.x =-2

x = (-2):7/2

x =-4/7

*-5/2.x+5=0

-5/2.x =0

x =0:-5/2

x =0

vậy x = -4/7 ; x=0