POTĘGI

1 OBILICZ

(0,1)²=
(²₃)⁴=
5¹-(¹₂)²=
(2¹₃)=
4²=
9⁰=
(-4)²=
-6²=
-7⁰=

2 0blicz stosując wzory

(3)²÷(1¹₂)²=
5²³×(¹₅)²³=
(1¹₂)²×(1²₃)²=
2⁴×5⁴=
3⁴×6⁴×4⁴=

ZApisz w postaci jednej potegi⁷
X⁸×X⁷=
X¹⁵×X³=
(6⁴)⁷=
16×4⁵×4=
y³⁰÷y¹⁷=
X¹²×X⁵

1

Odpowiedzi

2010-01-11T17:46:09+01:00
Zad 1

(0,1)²=0,1*0,1=0,01
(²₃)⁴=16/81
5¹-(¹₂)²=5-1/4=4i3/4
(2¹₃)=1/2^3=1/8
4²=4*4=8
9⁰=1
(-4)²=(-4)*(-4)=16
-6²=(-6)*(-6)=36
-7⁰=1

zad 2

(3)²÷(1¹₂)²=(3:1i1/2)^2=2^2=4
5²³×(¹₅)²³=(5*1/5)^23=1^23=1
(1¹₂)²×(1²₃)²=(1i1/2*1i2/3)^2=(5/2)^2=25/4=6i1/4
2⁴×5⁴=(2*5)^4=10^4=10000
3⁴×6⁴×4⁴=(3*6*4)^4=72^4=5184*5184=26873856

zad 3

X⁸×X⁷=x^15
X¹⁵×X³=x^18
(6⁴)⁷=6^28
16×4⁵×4=4^2*4^5*4^1=4^8
y³⁰÷y¹⁷=y^13
X¹²×X⁵=x^17
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