Tillbaka
A — 0901
1 p
Om
a
≥
0
a \ge 0
a
≥
0
,
b
≥
0
b \ge 0
b
≥
0
,
c
≥
0
c \ge 0
c
≥
0
, och
x
=
a
b
3
⋅
a
2
b
3
4
⋅
b
c
2
6
x = \sqrt{ab^3} \cdot \sqrt[4]{a^2b^3} \cdot \sqrt[6]{bc^2}
x
=
a
b
3
⋅
4
a
2
b
3
⋅
6
b
c
2
, så gäller
A
x
=
a
b
2
c
x = ab^2\sqrt{c}
x
=
a
b
2
c
B
x
=
a
b
2
b
c
x = ab^2\sqrt{bc}
x
=
a
b
2
b
c
C
x
=
a
b
b
c
x = ab\sqrt{bc}
x
=
ab
b
c
D
inget av (a)-(c).
Svara