X=π/4+πn, n∈Z [5π;13π/2]= [5π; 6,5π]
n=0 x1=π/4+0π=π/4∉ [5π; 6,5π]
n=1 x2=π/4+π=5π/4∉ [5π; 6,5π]
n=2 x3=π/4+2π=9π/4∉ [5π; 6,5π]
n=3 x4=π/4+3π=13π/4∉ [5π; 6,5π]
n=4 x5=π/4+4π=17π/4∉ [5π; 6,5π]
<u>n=5 x6=π/4+5π=21π/4∈[5π; 6,5π]</u>
<u>n=6 x7=π/4+7π=25π/4∈ [5π; 6,5π]
</u>n=7 x8=π/4+8π=33π/4∉ [5π; 6,5π]
Ответ: 21π/4; 25π/4
(x^5+2x^4-3x^3+2x^2-3x )=(x^2+x+1)(x^3+x^2-5x+6)-4x-6
Ответ на фото
<span>1) 2,4*4 7/12=11
2) </span><span>2,25:1 1/8=2
3) 11-2=9</span>