\(x^3-12x^2-28x=0\)
\(\Leftrightarrow x\left(x-14\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=14\\x=-2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử:
x2 - 2x - 4y2 + 1
3x2 - 12x + 12
12x2y5 - 32x3y4z + 28x4y3
x4 + 64
x2 - 2xz + 2yz - y2
tìm x:
a, 5x(12x-7)-6(10x^2+3)=0
b, 9x(3x+1-2x^2+4x-3)=0
Tìm x:
x3+6x2+12x+35=0
Tìm x
1, x3-x(x2+2) =6
2, (x-2)2-3(x-2) =0
3, x2+2x+1=0
4, 4x2-12x=-9
Tìm x, biết:
1) \(x^2-6x=0\)
2) \(2x^3-5x^2-12x=0\)
3) \(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)
Tìm x
x3-6x2+12x-35=0
tìm x biết
a)x^3-6x^2+12x-8-x(x^2-4x+4)=0
b)x(x^2+2)=3x+6
1, tìm x biết
a, (2x-1)^2-2x+1=0
b, (4x-3)^2-12x+9=0
c, 3x^2-48=0
Tìm x, biết
a) 3x3 - 6x2 = 0
b) x ( x-4) - 12x + 48 =0
c) x (x-4)- (x2-8) =0
d) 2x (x-5 ) - x (2x+3) =16
e) (4x2-1) - (x-1)2 = -3
