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
Ответ на задание под номером девять:
...= 4х(х-5)*1/(х-5)^2 = 4х*1/х-5 = 4х/х-5
{3x-9y=12|*4
{4x-12y=16|*3
-{12x-36y=48
-{12x-36y=48
0y=0
любой корень
допустим 1
3x-9*1=12
3x=12+9
3x=21:3
x=7