решить 1) (6x + 1) – (3 – 2x) = 14 2) (6 – 2c) + 4 = – 5c – 3 3) (10x–3) + (14x–4) = 8 – (15 – 22x) - id1184841320201111 от znasta745 11.11.2020 01:04

Question

Математика: решить 1) (6x + 1) – (3 – 2x) = 14 2) (6 – 2c) + 4 = – 5c – 3 3) (10x–3) + (14x–4) = 8 – (15 – 22x) 4) (7–10k) – (8–8k) + (10k + 6) = – 8 5) 6y – 20 = 2(5y – 10) – 4y 6) (1 + 5v) = 5(1 + 6v) 7) 3,4 – 0,6b= 2b – (0,4b+ 1) 8) – 3(y + 2,5) = 6,9 – 4,2y 9) 4(1 – 0,5a) = – 2(3 + 2a) 10) 3(0,4x + 7) – 4(0,8x – 3) = 2

znasta745 · Accepted Answer

найдите количество точек экстремума функции g(x)=x^7-35x^5

1)g'(x)=7x^6-35·5x^4=7x^4(x^2-25)

2)g'(x)=0 7x^4(x^2-25)=0 ⇔x1=0 x2=-5 x3=5

+ (g'(x)>0) - (g'(x)<0) - + (g'(x)>0)
3)g'(x)>0 (g'(x)<0) (-5)05
x2 x1 x3
max min

x2, x3 - точки экстремумов функции g(x)

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