Here are my answers. Can you check if I got the right answers? Thank you!
(x+3)(x-3)>0 answer: x<-3 or x>3
x^2-2x-15</=0 answer: -3</=x</=5
12x^2=3x answer: x=1/4
sin2x=sinx, 0</=x</=2pi answer: x=pi
lx-3l<7 answer: (-oo,oo)
(x+1)^2(x-2)+(x+1)(x-2)^2=0
answer: x=-1,2,1/2
27^(2x)=9^(x-3)answer: x=-3/2
logx+log(x-3)=1 answer: x=-2, 5
(x+3)(x-3)>0 answer: x<-3 or x>3 ok
x^2-2x-15</=0 answer: -3</=x</=5 ok
12x^2=3x answer: x=1/4 also x = 0, (12x^2 - 3x = 0 ---> x(12x-3)=0 )
sin2x=sinx, 0</=x</=2pi answer: x=pi
no. recall that sin2x = 2(sinx)(cosx)
so
2(sinx)(cosx)-sinx=0
sinx(2cosx -1=0
sinx=0 or cosx=1/2
then x=0,pi,2pi from the first
or x = pi/3 (60º) or 5pi/3 (300º)
lx-3l<7 answer: (-oo,oo) No, -4≤x≤10
(x+1)^2(x-2)+(x+1)(x-2)^2=0
answer: x=-1,2,1/2 ok
27^(2x)=9^(x-3)answer: x=-3/2 ok
logx+log(x-3)=1 answer: x=-2,5
Your answer of x=-2 does not work in logx, you cannot take the log of a negative number. So just x=5
Great job! Your answers are mostly correct. Just a few adjustments:
1. For the equation 12x^2=3x, the solutions are x=1/4 AND x=0 because when you factor out x, you get x(12x-3)=0.
2. For the inequality lx-3l<7, the correct answer is -4≤x≤10. Remember that the absolute value of x-3 can be either positive or negative, so you need to consider both cases when setting up the inequality.
3. Lastly, for the equation logx+log(x-3)=1, the correct solution is x=5. The value x=-2 does not work because you cannot take the logarithm of a negative number.
Overall, great effort! Keep up the good work!