Question

Rezolvati ecuatia |2x+3| * |x-1| + $\sqrt{4 x^{2}+12x+9 }$ * |2x-2| = 0 . Va rog mult!!!!

Answer 1

I2x+3I*Ix-1I+√(2x+3)^2* 2*Ix-1I=0
I2x+3I*Ix-1I+I2x+3I*2*Ix-1I=0
3I2x+3I*Ix-1I=0
I2x+3I=0,    2x+3=0, x=-3/2
sau Ix-1I=0,  x-1=0, x=1

Answer 2

4x²+12x+9=0
Δ=144-144=0
x1=x2=-12/8=-3/2
deci x²+12x+9=(2x+3)²

deci ex.va fi: I2x+3I·Ix-1I+[√(2x+3)²]·I2x-2I=0⇒
 I2x+3I·Ix-1I+(2x+3)·I2x-2I=0
2x+3=0⇒   2x=-3   ⇒x=-3/2
x-1=0⇒     x=1

x               -α                   -3/2                 1                     +α
2x+3          - - - - - - - - - - -  0 +++++++++++++++++++++++++++
x-1            - - - - - - - - - - - - - - - - - - - - - - -0+++++++++++++++++
                ++++++++++++0- - - - - - - - -  -0+++++++++++++++++
cazul I daca x∈(-α,-3/2]u[1,+α)=D1
(2x+3)((x-1)+(2x+3)·2(x-1)=0
(2x+3)(x-1)·(1+2)=0
3(2x+3)(x-1)=0
2x²-2x+3x-3=0
2x²+x-3=0
Δ=1+24=25
x1=(-1+5)/4=4/4=1∈D1
x2=(-1-5)/4=-6/4=-3/2∈D1
S1={-3/2,1}

cazul II daca x∈(-3/2,1)=D2
(-2x-3)(1-x)+(2x+3)(2-2x)=0
-2x+2x²-3+3x+4x-4x²+6-6x=0
-2x²+x+3=0     inmultim -1
2x²-x-3=0
Δ=1+24=25
x1=(1-5)/4=-4/4=-1∈D2
x2=(1+5)/4=6/4=3/2∉D2
S2={-1}
S=S1US2
S={-3/2,1}U{-1}
S={-3/2,-1,1}