ОДЗ
sinx>0⇒x∈(2πk;π+2πk,k∈z)
при любом х
2cosx=√2
cosx=√2/2
x=-π/4+2πk,k∈z не удов усл
x=π/4+2πk,k∈z
7π/2≤π/4+2πk≤5π
14≤1+8k≤20
13≤8k≤19
13/8≤k≤19/8
k=2⇒x=π/4+4π=17π/4
5! = 1*2*3*4*5=120
6! = 1*2*3*4*5*6=720
<span>5!+6!+7! / 8!-7! = 5!(1+6+6*7)/7!(8-1)=(7+42)/6*7*7=49/(42*7)=1/6
</span>
<span>|x|=5
</span>х=5 или х=-5
<span>|x|=10
</span>х=10 или х=-10
= - m + 2n + 3a + b = 3a + 2n + b - m
= х - 5 - 7 + х + 9 - х = (х + х - х) + (9 - 7 - 5) = х + (-3) = х - 3