Question

Am nevoie de ajutor ! dau coroana​

Answer 1

Răspuns:

Explicație pas cu pas:

1) z = a+bi,  z barat a -bi

A = (a+bi)(2+3i) + (a-bi)(2-3i) =

2a +3ai +2bi -3b + 2a -3i -2bi -3b =

4a - 6b  este in R

2) x^2 -6x + 7 = 0,   delta = 36-28 = 8

x1,2 = (6 +- 2√2)/2

x1 = 3 -√2,   x2 = 3 +√2

f(3 +√2) = 0 (e radacina), deci produsul = 0

3) x^2 +x -2 >0,  (x+2)(x-1) > 0

Radacini:  -2  si 1

In afara rad. semn diferit de a, x^2 +x -2 <0

Asadar x^2 +x -2 >0 pt.  x in afara rad. ,

adica x in (-inf, -2)U(1, +inf)

(x-1)/2 > 0,   x > 1

Deci cautam radacini in (1, +inf)

lg(x^2 +x -2) = lg10 + lg((x-1)/2)

lg(x^2 +x -2 ) = lg(10*(x-1)/2) = lg(5(x-1))

x^2 +x -2 = 5x -5

x^2 -4x + 3 = 0,   delta = 16 -12 = 4

x1,2 = (4 +- 2)/2

x1 = 2/2 = 1  nu convine

x2 = 6/2 = 3  convine

Cam atat...