a, (3x^3-5x^2+9x-15):(3x-5) b, x^3-3x^2 c, x^2-2016-2017 d, 3(x+1)-2x+1=0 e, x^2-4x -5=0 02/08/2021 Bởi Claire a, (3x^3-5x^2+9x-15):(3x-5) b, x^3-3x^2 c, x^2-2016-2017 d, 3(x+1)-2x+1=0 e, x^2-4x -5=0
a) (3x³ – 5x² + 9x – 15) : (3x – 5) = [x²(3x – 5) + 3(3x – 5)] : (3x – 5) = [(3x – 5)(x² + 3)] : (3x – 5) = x² + 3 b) x³ – 3x² = x² (x – 3) c) x² – 2016 – 2017 = x² – 4033 d) 3(x + 1) – 2x + 1 = 0 ⇔ 3x + 3 – 2x + 1 = 0 ⇔ x + 4 = 0 ⇔ x = -4 Vậy x = -4 e) x² – 4x – 5 = 0 ⇔ x² + x – 5x – 5 = 0 ⇔ x(x + 1) – 5(x + 1) = 0 ⇔ (x + 1)(x – 5) = 0 ⇔ \(\left[ \begin{array}{l}x+1=0\\x-5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\) Vậy $x=1$ hoặc $x=5$ Bình luận
(3x³-5x²+9x-15):(3x-5) =[x²(3x-5)+3(3x-5)]:(3x-5) =(3x-5)(x²+3):(3x-5) =x²+3 (Học tốt nhé!) Bình luận
a) (3x³ – 5x² + 9x – 15) : (3x – 5)
= [x²(3x – 5) + 3(3x – 5)] : (3x – 5)
= [(3x – 5)(x² + 3)] : (3x – 5)
= x² + 3
b) x³ – 3x²
= x² (x – 3)
c) x² – 2016 – 2017
= x² – 4033
d) 3(x + 1) – 2x + 1 = 0
⇔ 3x + 3 – 2x + 1 = 0
⇔ x + 4 = 0
⇔ x = -4
Vậy x = -4
e) x² – 4x – 5 = 0
⇔ x² + x – 5x – 5 = 0
⇔ x(x + 1) – 5(x + 1) = 0
⇔ (x + 1)(x – 5) = 0
⇔ \(\left[ \begin{array}{l}x+1=0\\x-5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
Vậy $x=1$ hoặc $x=5$
(3x³-5x²+9x-15):(3x-5)
=[x²(3x-5)+3(3x-5)]:(3x-5)
=(3x-5)(x²+3):(3x-5)
=x²+3
(Học tốt nhé!)