Y`=8x+7; y`(8)=64+7=71; y(8)=4*64+56+2=314
касательная
f(x)=y(x0)+f`(x0)(x-x0)=314+71(x-8)
нормаль
g(x0)=y(x0)-(x-x0)/y`(x0)=314-(x-8)/71
X²-x≥6
x≥-1
x²-x-6≥0
D=1-4*1*(-6)=25
x(1)≥(1-√25)/2 x(1)≥-2
x(2)≥(1+√25)/2 x(2)≥3
Ответ: x≥3
5tgx - 12/tgx + 11 = 0
5tg²x + 11tgx - 12 = 0 (tgx ≠ 0)
Пусть t = tgx.
5t² + 11t - 12 = 0
D = 121 + 5•12•4 = 361 = 19²
t1 = (-11 + 19)/10 = 8/10 = 4/5
t2 = (-11 - 19)/10 = -30/10 = -3
Обратная замена:
tgx = 4/5
x = arctg(4/5) + πn, n ∈ Z
tgx = -3
x = arctg(-3) + πn, n ∈ Z.
1. a) ...=b(2a-3)-2a(b-2)/2ab=2ab-3b-2ab+4a=-3b+4a
б) ...=(a+3)×3(a+3)×96a^4/8a(a+3)=3(a+3)×12a^3=3a+9×12a^3
в) ...=(x-y)(x+y)/2x(x+y)=x-y/2x