急!!! CALCULUS問題!!

Q.1
xlnx - 3 + lnx -3x =0
xlnx + lnx = 3x +3 +3(x +1)
lnx (x +1) = 3(x +1)
therefore, lnx = 3 , x =e^3.
Q.2
x^3 - x^(5/2) -6x^2 = 0
x^2(x -sqrtx -6) =0
x^2 (sqrtx -3)(sqrtx + 2) = 0
Therefore, x =0, 4 or 9.

2008-06-23 08:26:42 補充：
Correction: Q.1 line 2 should be 3x + 3 =3(x +1).

2008-06-23 08:30:55 補充：
x=4 is rejected because it does not fit the equation. So answers are 0 and 9 only.

xlnx− 3+lnx−3x=0
xlnx+lnx-(3x+3)=0
lnx(x+1)-3(x+1)=0
(x+1)(lnx-3)=0
x=-1(rej.) or lnx=3
∴x=e3
x3−x5/2−6x2=0
x2(x-x1/2-6)=0
x2(x1/2-3)(x1/2+2)=0
x2=0 or x1/2=3 or -2
x=0, 9 or 4(rej.)