Question

Puteți rezolva testul 2, va rog?

Answer 1

Răspuns:

Explicație pas cu pas:

T2 1a) 2^(x+√x) = 2^(2x -2√x)

x +√x = 2x -2√x

x -3√x = 0, √x*√x = 3√x

√x = 0,  x = 0 si   √x = 3,  x = 9

b) 1/(2^x+5) + 10/(2^x +8) = 1

y = 2^x +5

1/y + 10/(y +3) = 1

y+3 +10y = y(y+3)

y^2 -8y -3 = 0, delta = 64 +12 = 76

?? 76 nu e patrat perfect,  se complica mult...

2) 2^(2x)/4 -2*2^x +3 = 0

y = 2^x

y^2 -8y +3 = 0,  delta = 64 -48 = 16

y1,2 = (8 -+4)/2,  y1 = 4/2 = 2,  y2 = 12/2 = 6

2^x = 2 ,   si 2^x = 6

Pt. x=1 :

|2 -2| +3|4-4| = m,   m = 0

Pt. 2^x = 6:

|6 -2) +|6^2 -4|  = m,  m = 44

3) 2^x +8 > 0 pt. x in R

10 -2^x >= 0,  10 > 2^x,  0 <= x <= 3

se vede ca x = 0  este solutie,  singura