Toán Lớp 8: Tìm x :
a) (x – 1)(2x + 3) – x(x – 1) = 0; b) x2 – 4x + 8 = 2x – 1.
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: Tìm x :
a) (x – 1)(2x + 3) – x(x – 1) = 0; b) x2 – 4x + 8 = 2x – 1.
Comments ( 2 )
a) ( x – 1 ) . ( 2x + 3 ) – x . ( x – 1 ) = 0
⇔ ( x – 1 ) . ( 2x + 3 – x ) = 0
⇔ ( x – 1 ) . ( x + 3 ) = 0
⇔ \(\left[ \begin{array}{l}x – 1 = 0\\x + 3 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 1\\x = – 3\end{array} \right.\)
Vậy , x ∈ { 1 ; – 3 }
b) x^2 – 4x + 8 = 2x – 1
⇔ x^2 – 4x + 8 – 2x + 1 = 0
⇔ x^2 – 6x + 9 = 0
⇔ x^2 – 2 . 3 . x + 3^2 = 0
⇔ ( x – 3 )^2 = 0
⇔ x – 3 = 0
⇔ x = 3
Vậy , x = 3 .
a) (x-1)(2x+3)-x(x-1)=0
⇔(x-1)(2x+3-x)=0
⇔(x-1)(x+3)=0
$⇔\left[\begin{matrix} x-1=0\\ x+3=0 \end{matrix}\right.$
$⇔\left[\begin{matrix} x=1\\ x=-3\end{matrix}\right.$
Vậy x∈{-3;1}
b) x^2 -4x+8=2x-1
⇔x^2 -4x-2x+8+1=0
⇔x^2 -6x+9=0
⇔(x-3)^2=0
⇔x-3=0
⇔x=3
Vậy x∈{3}