Решение
<span>sin х, если cos х= 8/17; -π/2<х<0
</span>sinx = √(1 - cos²x) = √(1 - (8/17)²) = √(1 - 64/289) = √(225/289) = 15/17
1)(x²-9)/(16x²-8x+1)>0
(x-3)(x+3)/(4x-1)²>0
(4x-1)²>0;=>x€(-oo;1/4)U(1/4;+oo)
(x-3)(x+3)>0;=>x€(-oo;-3)U(3;+oo)
__-3_______1/4________3____
=>x€(-oo;-3)U(3;+oo)
2){3-x>0;x<3
{3-x≠1;x≠4
=>x<3
3){x<3
{x€(-oo;-3)U(3;+oo);=>x€(-oo;-3)
Z²+13z+30=0
мо теореме Виета:
z1 + z2 = -13
z1z2 = 30
z1 = -3
z2 = -10
Ответ: -10; -3.
X²-2x-35=0
x1+x2=2 U x1*x2=-35
x1=7 U x2=-5
2/(x-7)+5/(x+5)-24/(x-7)(x+5)=1
x≠7 u x≠-5
2(x+5)+5(x-7)=x²-2x-35
x²-2x-35-2x-10-5x+35+24=0
x²-9x+14=0
x1+x2=8 U x1*x2=14
x1=2
x2=7- не удов усл
Ответ х=2