Question

Zadania które muszę oddać po feriach. Z różnych działów.

Potęgi.

Równania i nierówności.

Zad. 1 Rozwiąż.

Zad. 2 Rozwiąż.

Answer 1

1

(-2)^6+(3^2-2^5)=64+9-32=41

(1/3)^2+(1/3)^1=1/6 + 1/3=1/6 +2/6=3/6=1/3

(-2)^6+(3^2-2^5)=64+9-32=41

(5-7)^2+(5-4)^2=(25-70+49)+1=4+1=5

(-3)^4:3^2-(-4)^3*(1/2)^4=81:9-(-64)*(1/16)=9+4=13

(-1  1/2)^2:(-3/4)^2+2  1/2*(-1/2)^5=(-3/2)^2:(-3/4)^2+5/2*(-1/2)^5=(-9/4):9/16+5/2*(-1/32)=-2-5/64=-133/64

2

6/x=2/3  /:6          x=9

x/12=4/8   /*12      x=6

Answer 2

Zad1

(-2)⁶ + (3² - 2⁵) = 64 + 9 - 32 = 41

(-⅓)² + (⅓)¹ = ⅙ + ⅓ = ⅙ + 2/6 = ½

(-2)⁶ + (3² - 2⁵) = 64 + 9 - 32 = 41

(5-7)² + (5-4)⁵ = (-2)² + 1⁵ = 4 + 1 = 5

(-3)⁴ : 3² -(-4)³ * (½)⁴ = 81 : 9 - (-64) * 1/16 = 9 + 4 = 13

(- 1½)² : (-¾)² + 2½ * (-½)⁵ = 2¼ : 9/16 + 2½ * (-1/32) = 4 - 5/64 = 3 59/64

 

Zad2

 

6/x = ⅔ /*x

⅔x = 6 /: ⅔

x= 9

 

x/12 = 4/8 /*12

x = 4/8 * 12

x= 6