task/30214294 решить уравнение : (2x²+5x+3) / (2x+3)=x² - x - 2.
решение (2x²+5x+3) / (2x+3)=x²-x-2 ⇔(2x+3)(x+1) / (2x+3) =x²-x-2
ОДЗ : 2x+3 ≠ 0
(2x+3)(x+1) / (2x+3) = x² - x - 2 ⇔ x + 1 = x² - x - 2 ⇔ x + 1 = (x + 1) (x - 2 ) ⇔ (x + 1 ) - (x + 1) (x - 2 ) =0 ⇔(x + 1 )(1 - (x - 2 ) ) =0 ⇔(x + 1 )(3 - x ) =0 ⇔
[ x + 1 =0 ; 3 - x = 0. ⇔ [ x = - 1 ; x = 3.
ответ: - 1 ; 3.
* * * ax²+bx +c = a(x -x₁)(x -x₂) * * *
2x²+5x+3 =0 ; D = 5² - 4*2*3 =25 -24 =1²
x₁,₂ =(-5 ± 1) /4 x₁ = (-5 -1) /4 = -3/2 , x₂ =(-5 +1) /4 = - 1
2x²+5x+3 =2( x - (-3/2) ) ( x- (-1) ) = (2x +3) ( x+1)
x² - x - 2 =0 ; D = 1² -4*1*(-2) =9 =3² ⇒
x₁ ,₂ =(1±3)/2 ⇒ x₁ = (1 -3) /2 = - 1 , x₂ = (1+3)/2 = 2 .
x² - x - 2 = (x -(-1) ) (x -2) = (x+1) (x -2)
task/30214294 решить уравнение : (2x²+5x+3) / (2x+3)=x² - x - 2.
решение (2x²+5x+3) / (2x+3)=x²-x-2 ⇔(2x+3)(x+1) / (2x+3) =x²-x-2
ОДЗ : 2x+3 ≠ 0
(2x+3)(x+1) / (2x+3) = x² - x - 2 ⇔ x + 1 = x² - x - 2 ⇔ x + 1 = (x + 1) (x - 2 ) ⇔ (x + 1 ) - (x + 1) (x - 2 ) =0 ⇔(x + 1 )(1 - (x - 2 ) ) =0 ⇔(x + 1 )(3 - x ) =0 ⇔
[ x + 1 =0 ; 3 - x = 0. ⇔ [ x = - 1 ; x = 3.
ответ: - 1 ; 3.
* * * ax²+bx +c = a(x -x₁)(x -x₂) * * *
2x²+5x+3 =0 ; D = 5² - 4*2*3 =25 -24 =1²
x₁,₂ =(-5 ± 1) /4 x₁ = (-5 -1) /4 = -3/2 , x₂ =(-5 +1) /4 = - 1
2x²+5x+3 =2( x - (-3/2) ) ( x- (-1) ) = (2x +3) ( x+1)
x² - x - 2 =0 ; D = 1² -4*1*(-2) =9 =3² ⇒
x₁ ,₂ =(1±3)/2 ⇒ x₁ = (1 -3) /2 = - 1 , x₂ = (1+3)/2 = 2 .
x² - x - 2 = (x -(-1) ) (x -2) = (x+1) (x -2)