Toán Lớp 8: Tìm x, biết 2x (x+3)-5 (x+3) = 0
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Toán Lớp 8: Tìm x, biết 2x (x+3)-5 (x+3) = 0
Comments ( 2 )
2x(x + 3) – 5(x + 3) = 0
<=> (x + 3)(2x – 5) = 0
=> \(\left[ \begin{array}{l}x + 3 = 0\\2x – 5 = 0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x = 0 – 3\\2x = 0 + 5\end{array} \right.\)
=> \(\left[ \begin{array}{l}x = -3\\2x = 5\end{array} \right.\)
=> \(\left[ \begin{array}{l}x = -3\\x = 5 : 2\end{array} \right.\)
=> $\left[\begin{matrix} x = -3\\ x = \dfrac{5}{2}\end{matrix}\right.$
Vậy: x ∈ {-3; 5/2}
2x.(x+3)-5.(x+3)=0
->(x+3).(2x-5)=0
->\(\left[ \begin{array}{l}x+3=0\\2x-5=0\end{array} \right.\)
->\(\left[ \begin{array}{l}x=-3\\2x=5\end{array} \right.\)
->\(\left[ \begin{array}{l}x=-3\\x=\dfrac{5}{2}\end{array} \right.\)
Vậy x in {-3; \frac{5}{2}}