\(x^2\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)
Tìm x : a) x^3 - 1/4x = 0 b) ( 2x - 1 )^2 - ( x + 3 )^2=0 c) x^2(x-3)+12-4x=0
a.4x^3-4x^2+x=0
b.x.(x-3)+12-4x=0
c.x^3+3x^2+3x-7=0
*tìm x*
Tìm x
x^3 - 3x^2 - 4x + 12 = 0
x^4 + x^3 - 4x -4 = 0
Tìm x
x^2 - 4x - 12 = 0
4x^2 + 4x - 24 = 0
8x^3 - 12x^2 + 6x - 1 = 0
b) (2x−1)2−(x+3)2=0;
c) x2(x−3)+12−4x=0
.
Tím x
a) (x-3)2-25x2=0
b) x3-1/4x=0
c) x(x-1)-(x-2)(x+2)=13
d) x2(x-3)+12-4x=0
e, x2-4+(x-2)2 = 0
f, x3+1=x(x+1)
g, x2(x-3)+12-4x = 0
h, (2x-1)2-(x+3)2 = 0
b) x ( x - 2 ) + x - 2 = 0
c) 5x ( x - 3 ) - x + 3 = 0
d) x2 ( x - 3 ) + 12 - 4x = 0
