Если по клеточкам строить, то всё должно красиво получиться
<span>a) y=log5(x^2-5x+6), x^2-5x+6>0, x^2-5x+6=0 при х=2 х=3
2 3
____+____I_____-_______I______+______>
D(f)=(- ∞,2)U(3,+∞)
б)y=log2/3(-x^2-5x+14), </span><span>-x^2-5x+14>0, </span><span><span>-x^2-5x+14=0 при х=- 7, х=2
- -7 + 2 -
______I____________I______________></span>
D(f)=(-7, 2)
в)y=log9(x^2-13x+12), x^2-13x+12>0, </span><span><span>x^2-13x+12=0 при х=1, х=12
+ 1 - 12 +
______I_________I__________>
D(f)=(-∞,1)U(12,+∞)
</span>г)y=log0,2(-x^2+8x+9), </span><span>-x^2+8x+9>0, </span><span>-x^2+8x+9=0, x=-1, x=9
</span>
<span><span><span> - -1 + 9 -
______I____________I______________></span>
D(f)=(-1, 9)</span>
</span>
Log₀,₅ (2x+1)>-2⇔ { 2x+1>0 и log₀,₅ (2x+1)> log₀,₅ (0,5^(-2)) ⇔
{2x+1>0 { x>-1/2
{(2x+1)<4 ⇔ { x<3/2 -----------(-1/2)/////////////////////////(3/2)-----
x∈(-1/2;3/2)
(sinx+sin3x)-(sin2x+sin4x)=0
2sin2xcosx-2sin3xcosx=0
2cosx*(sin2x-sin3x)=0
cosx=0⇒x=π/2+πn,n∈z
sin2x-sin3x=0
-2sin(x/2)cos(5x/2)=0
sin(x/2)=0⇒x/2=πk⇒x=2πk,k∈z
cos(5x/2)=0⇒5x/2=π/2+πm⇒x=π/5+2πm/5,m∈z