f)
|x³+3x²+3x+1-(x³-6x²+12x-8)|=9
|9x²-9x+9|=9
9|x²-x+1|=9
|x²-x+1|=1
x²-x+1=1
x²-x=0
x(x-1)=0...x1=0....x2=1
x²-x+1=-1
x²-x+2=0 x3,4∉R..nu stim unde ne-a spus sa rezolvam ec are solutii in C
x3,4=(1+-i√7)/2
g)
C.E. x≠2
modulul produsului(deci si al raportului) este egal cu [produsul modulelor
|3|/|x³-8|=|1|/|x-2|
tinem cont ca x³-8=x³-2³=(x-2)(x²+2x+4)
deci
3/|x-2||x²+2x+4|=1/|x-2|
|x-2||x²+2x+4|/3=|x-2|/1
|x-2|(|x²+2x+4|/3-1)=0
dar |x-2|≠0, pt ca x≠2
ramane
|x²+2x+4|=3
x²+2x+4=3
x²+2x+1=0...
(x+1)²=0......x1=x2=-1∈R
x²+2x+4=-3
x²+2x+7=0 x3,4∉R..x3,4=(-2+-2√6*i)/2=(-1+-i√6)∈C
3|x-2||x²
