Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(5x^2\) - 3x =0
x(5x - 3) =0
TH1:x=0
TH2:
5x-3=0
5x=3
x=\(\dfrac{3}{5}\)
Vậy x=0 ; x= \(\dfrac{3}{5}\)
5x2 - 3x = 0
<=> x(5x - 3) = 0
<=> x = 0 hoặc 5x - 3 = 0 <=> x = 3/5
Vay x = 0 hoặc x = 3/5
1. (4x-10).(24+5x)=0
2 .(2x-5).(3x-2)=0
3. (2x-1).(3x+1)=0
4. x.(\(x\)2-1)=0
5.(5x+3).(\(x^2\)+4).(x-1)=0
6.(x-1).(x+2).(x+3)=0
7.(x-1).(x+5).(-3x+8)=0
Bài 1: Tìm x
a) ( 2x-1)2 - (x + 3)2 = 0
b) 5x. ( x-3) + 3 - x = 0
c) ( 5x2 + 3x - 2) 2 = ( 4x2 - 3x -2)
d) x3 + 27 + (x+3) . ( x-9) = 0
6x - 8y/x+y+3y-2x/5x-y biết 3x2-65xy+16y2=0. y#-x,y#5x
TÌm x:
( x + 2)4 + (x -4) 4 = 162
x2 - 5x - 21 = 0
x2 - (x+1)2 = 0
(x - 1).(x+2) - x-2= 0
x2 - 4x +3 = 0
3x ( x-4) +12x - 48 = 0
(2x-1).(5-3x) = (x+2).(5-3x)
chứng minh các phương trình sau vô nghiệm:
a, x4 - 3x2 + 6x + 13 = 0
b, 3x4 - 5x3 + 8x2 - 5x + 3 = 0
(x2-3x+2)(x2+15x+56)+8=0
(2x2-3x+1)(2x2-5x+1)-9x2=0
Tìm x biết:
\(a,3x^3-3x=0\)
\(b,x\left(x-2\right)+x-2=0\)
\(c,5x\left(x-2000\right)-x+2000=0\)
Tìm x biết:
4x2 - 6x = 0
b) 4x2 + 4x = -1
c) 5x2 + x = 0
d) x3 - 5x = 4x2
3x(x-2) = x-2
x3 - 16x = 0
Tìm x
2) 2 (x + 5) - x\(^2\) - 5x = 0
3) x\(^2\) + 3x + 2 = 0
4) x\(^2\) - 4x - 5 = 0
5) -2x\(^2\) - 3x + 5 = 0
