a 2x . ( x + 3 ) – 2x^2 + 1 = 0
b ( x + 2 ) . ( x – 2 ) – x . ( x + 5 ) = 6
a 2x . ( x + 3 ) – 2x^2 + 1 = 0 b ( x + 2 ) . ( x – 2 ) – x . ( x + 5 ) = 6
By Lydia
By Lydia
a 2x . ( x + 3 ) – 2x^2 + 1 = 0
b ( x + 2 ) . ( x – 2 ) – x . ( x + 5 ) = 6
Đáp án:
ở dưới
Giải thích các bước giải:
$a)⇔2x^2+6x-2x^2+1=0$
$⇔6x=-1$
$⇔x=-1/6$
$b)⇔x^2-2x+2x-4-x^2-5x=6$
$⇔-5x=10$
$⇔x=-2$
a 2x . ( x + 3 ) – 2x^2 + 1 = 0
<=> 2x^2+6x-2x^2+1=0
6x+1=0
6x=-1
x=-1/6
b ( x + 2 ) . ( x – 2 ) – x . ( x + 5 ) = 6
x^2-2x+2x-4-x^2-5x=6
-5x-4=6
-5x=10
x=-2