Question

5si 6 dau coroana ajutorrrrrr

Answer 1

1) arcsin(-rad2/2) = -pi/4 = tg(-pi/4) = -1    arccos(-1) = pi   =>  Raspuns:  pi - 1
2) x = arctg(a) + k*pi
3) Dreapta are panta                  m1 =-a/b adica -(-2)/5 = 2/5
  Conditia de perpend.  m1*m2 = -1 => m2 = -5/2
 Ec.perpend. ce trece prin A(-1,3) cu panta m2 :   y - 3 = -5/2(x-(-1))  =>
 2y - 6 = -5x - 5 =>  -5x - 2y +1 = 0
4) sin(2x + pi/3) = sin(-pi/3) => 2x+pi/3 = -pi/3 => 2x=2pi/3 => x=pi/3
5) Notam  y = 5^x  ; 5^(2x) =(5^x)^2  ; 5^(x+1) = 5*5^x 
 2*y^2 -5*y = 25 => 2*y^2 -5*y -25 = 0 => delta = 25 + 200 = 225  rad(225) = 15 =>  x1 = (5+15)/4 = 5 ;   x2 = (5-15)/4 = -10/4  Convine numai x1 => 5^x = 5 => x = 1
6) Se impune conditia 2x + 1 >= 0 => x >= -1/2  rad(2x+1) = 7 - x => ridicam la patrat: 2x+1 = 49 - 14x + x^2  x^2 - 16x + 48 =0 => delta = 256 - 192 = 64 
 x1 = (16+8)/2 = 12 ;     x2 = (16-8)/2 = 4
 Ambele sol. sunt bune
7) Se impune conditia x + 1 > 0 => x > -1  2 = log3(3^2)  log3(3^2) + log3(x+1) = log3(5x^2 + 7)  lo3(3^2 * (x + 1)) = log3(5x^2 + 7) => 9(x + 1) = 5x^2 + 7  5x^2 -9x - 2 = 0 => delta = 81 + 40 = 121  x1 = (9+11)/10 = 2 ;  x2 = (9-11)/10 = -1/5
  Ambele sol. sunt bune