(2x + 5)(4x2 - 9) = 0
<=> \(\left[{}\begin{matrix}2x+5=0\\4x^2-9=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x^2=\dfrac{9}{4}< =>x=\pm\dfrac{3}{2}\end{matrix}\right.\)
KL: \(x\in\left\{\dfrac{-5}{2};\pm\dfrac{3}{2}\right\}\)
Tìm x biết :
a, (x-2).(x2 +2x +7) +2.( x2-4) -5 .(x-2) =0
b, 4x2 -25 -(2x-5) .(2x+7) =0
c, x3 +27 + (x+3) .(x-9)=0
. Tìm x, biết:
a) 4x2 – 9 = 0
b) (x + 5)2 – (x – 1)2= 0
c) x2 – 6x – 7 = 0
d) (x + 1)2 – (2x - 1)2 = 0
Tìm x biết ( 2 x – 3 ) 2 – 4 x 2 + 9 = 0
A. x = 1 2
B. x = - 3 2
C. x = 3 2
D. x = 2 3
Giải các phương trình sau:
a) 9 − x = 2 x ; b) x − 15 + 1 = 3 x ;
c) 4 x 2 − 1 + 3 x 2 x − 1 = 0 ; d) 5 − 4 x = 4 − 5 x .
Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
(2x-3)2+4x2-9=(2x-3)(3x+5)
4x2-3(2x-5)-25=0
Bài 5. Tìm x, biết:
a) x (2x - 7) + 4x -14 = 0
b) x3 - 9x = 0
c) 4x2 -1 - 2(2x -1)2 = 0
d) (x3 - x2 ) - 4x2 + 8x - 4 = 0
Tìm X:
a) 16x2-24x+9=25
b) x2+10x+9=0
c) x2-4x-12=0
d) x2-5x-6=0
e) 4x2-3x-1=0
f) x4+4x2-5=0
