Log₁/₆(10-x)+log₁/₆(x-3)≥-1
ОДЗ: 10-x>0 x<10 x-3>0 x>3 x∈(3;10)
log₁/₆(10-x)(x-3)≥log₁/₆6
-x²+13x-30≤6 I×(-1)
x²-13x+36≥0 D=25
x₁=4 x₂=9
(x-4)(x-9)≥0
-∞_____+_____4_____-_____9_____+______+∞
x∈(-∞;4]U[9;+∞)
Учитывая ОДЗ х∈(3;4]U[9;10).
ОДЗ
x ≥ 0
( - 2x + 3)^2 = x^2
(-2x)^2 - 12x + 9 - x^2 = 0
4x^2 - 12x + 9 - x^2 = 0
3x^2 - 12x + 9 = 0 /:3
x^2 - 3x + 2 = 0
D = 9 - 8 = 1
x1 = ( 3 + 1)/2 = 4/2 = 2
x2 = (3 - 1)/2 = 2/2 = 1
Ответ
1; 2
D=b^2-4ac
D=1^2-4*6*(-1)=25
Sinx/cosx>cosx/cosx
tgx>1
π/4+πk<x<π/2+πk
Sin2x-√3cosx=0
2sinxcosx-√3cosx=0
cosx(2sinx-√3)=0
cosx=0⇒x=π/2+πn
sinx=√3/2⇒x=(-1)^n*π/3+πn
x=-7π/2;-23π/6;-11π/3