\(-2x^2+9x0\Leftrightarrow x\left(2x-9\right)>0\)
\(\Leftrightarrow\left(x0\right)\)
\(\Leftrightarrow\left(x\frac{9}{2}\right)\)
\(\Leftrightarrow x\frac{9}{2}\)
Vậy x < 0 hoặc x > 9/2.
bài 2; tìm x
a, 5x ( x - 1 ) + ( x + 17 ) = 0
b, 3x ( x - 3 ) mũ 2 - 3x ( x + 3 ) mũ 2 = 0
c, 7 - 9x + 2x mũ 2 = 0
d, 7 - 9x + 2x mũ 2 = 0
2x4-9x3+14x2-9x+2=0
a,(2x-5)^2-x^2=0
b.(x^2-2x+1)-9x^2=0
a,(2x-5)^2-x^2=0
b.(x^2-2x+1)-9x^2=0
Tìm x:
x^2 + 6x
x^2 - 25x + 250 = 0
x^2 + 9x = 10
2x^2 + 9x = 35
(x^2 - 2x - 1)^2 - 5 (x^2 - 2x - 1) - 14 = 0
(2k^2 + 5k + 1)^2 - 12 (2k^2 + 5k + 1) + 32 = 0
Giải phương trình:2x^4-7x^3+9x^2-7x+2x^2=0
Tìm x
a) (2x - 3)(x^2 + 2) - 2(x + 1)^3 - 9x^2 = -5
b) 3(x - 2) - x^2 + 4 = 0
c) x^3 - 5x^2 - 10x= -50
d) x^3 + 9x= 6x^2
e) 2x^2 - 5x + 3 = 0
f) x^2 - x - 2= 0
Tìm x:
a)9x^2-30x+25=0
b)25x^2-5x+1/4=0
c)9x^2-25=0
d)(2x-1)^2-(3x+2)^2=0
a) (x – 2)(x2 + 2x + 4) – x( x2 +2) = 12 b) (x – 3)2 – (x+2)(x–2) = 16
c) x3 – 9x = 0 d) x3 – 6x2 + 9x – 54 = 0
giúp e vs ạ
