Toán Lớp 8: Tìm x:
a)2x(x+3)-4x-12=0
b)9x²-25=0
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Toán Lớp 8: Tìm x:
a)2x(x+3)-4x-12=0
b)9x²-25=0
Comments ( 2 )
Giải đáp+Lời giải và giải thích chi tiết:
a)
2x(x+3)-4x-12=0
⇔2x(x+3)-(4x+12)=0
⇔2x(x+3)-4(x+3)=0
⇔(x+3)(2x-4)=0
$⇔\left[\begin{matrix} x+3=0\\ 2x-4=0\end{matrix}\right.$
$⇔\left[\begin{matrix} x=-3\\ 2x=4\end{matrix}\right.$
$⇔\left[\begin{matrix} x=-3\\ x=2\end{matrix}\right.$
Vậy x=-3 hoặc x=2
b)
9x^2-25=0
⇔(3x)^2-5^2=0
⇔(3x-5)(3x+5)=0
$⇔\left[\begin{matrix} 3x-5=0\\ 3x+5=0\end{matrix}\right.$
$⇔\left[\begin{matrix} 3x=5\\ 3x=-5\end{matrix}\right.$
$⇔\left[\begin{matrix} x=\dfrac{5}{3}\\ x=- \dfrac{5}{3}\end{matrix}\right.$
Vậy x=5/3 hoặc x=- 5/3
->Áp dụng HĐT: A^2-B^2=(A-B)(A+B)
#Linhsad
a) 2x(x+3)-4x-12=0
⇔ 2x(x+3)-4(x+3)=0
⇔ (x+3)(2x-4)=0
⇔ (x+3)(x-2)=0
⇔ $\left[\begin{matrix} x+3=0\\ x-2=0\end{matrix}\right.$
⇔ $\left[\begin{matrix} x=-3\\ x=2\end{matrix}\right.$
Vậy x∈{-3,2}
b) 9x²-25=0
⇔ (3x-5)(3x+5)=0
⇔ $\left[\begin{matrix} 3x-5=0\\ 3x+5=0\end{matrix}\right.$
⇔ $\left[\begin{matrix} 3x=5\\ 3x=-5\end{matrix}\right.$
⇔ $\left[\begin{matrix} x=\dfrac{5}{3}\\ x=\dfrac{-5}{3}\end{matrix}\right.$
Vậy x∈{5/3;-5/3}