SKEMA – JORDBUNDSANALYSE

1
x kg
2 kg 2 kg 2 kg 2 kg
x kg
x kg
x kg
2 kg 2 kg 2 kg 2 kg 2 kg 2 kg
5
Hvert lod vejer ·________
kg.
5
4x = 2 · 10
Skriv ligningen: ······················································ Løsningen
er: ······················································
2 · 10 = 4x ⇔························································
4x = 20 ⇔
Forklar hvordan ligningen skal løses: ························································
20
x= 4 ⇔x=5
···········································································································································································
2
x kg
x kg
2 kg
x kg
2 kg 2 kg 2 kg 2 kg 2 kg
4
Hvert lod vejer ···············
kg.
4
3x = 6 · 2
Skriv ligningen: ······················································ Løsningen
er: ······················································
3x = 6 · 2 ⇔···························································
3x = 12 ⇔
Forklar hvordan ligningen skal løses: ·····················································
12
x= 3 ⇔x=4
···········································································································································································
3
10 kg
x kg
x kg
10 kg
10 kg
10 kg
10 kg
25 kg.
Hvert lod vejer ···············
2x = 10 · 5
Skriv ligningen: ······················································ 25
Løsningen er: ······················································
2x = 10 · 5 ⇔ 2x = 50 ⇔
Forklar hvordan ligningen skal løses: ···············································································································
50
x = 2 ⇔ x = 25
···········································································································································································
M AT E M AT R I X 7
15
ALINEA
LIGNINGER
1
Forklar hvad der er sket med ligningerne fra linje til linje og find løsningen.
a
Ligning
Forklaring
3x + 2 = 14 3x + 2 – 2 = 14 – 2
Der er trukket 2 fra
på begge sider af lighedstegnet
3x = 12
Der er reduceret
på begge sider af lighedstegnet
Der er divideret med 3
på begge sider af lighedstegnet
3x
3
=
12
3
x=4
Løsningen er 4
b
Ligning
Forklaring
8x – 4 – 2x + 7 = 21
6x + 3 = 21
Der er reduceret på venstre side.
6x + 3 – 3 = 21 – 3
Der er trukket 3 fra på begge sider.
6x = 18
Der er reduceret på begge sider.
6x
6
=
18
6
Der er divideret med 6 på begge sider.
3
x = ···················
Løsningen er
3
c
Ligning
Forklaring
40 + 2x – 3x + 5x = 36
40 + 4x = 36
Der er reduceret på venstre side.
40 + 4x –40 = 36 – 40
Der er trukket 40 fra på begge sider.
4x = –4
Der er reduceret på begge sider.
4x
4
=
–4
4
Der er divideret med 4 på begge sider.
–1
x = ···················
Løsningen er
2
Find den ubekendte.
a
2a = 8
a=
b
3b +1 = 10
b=
4
c
12c – 5 = 19
c=
d 2d
3
M AT E M AT R I X 7
–1
2
=8
1
4
d=
16
ALINEA
e
2e – 12 e = 5
e=
f
2f + 0,5 = 1,5
f=
3 13
1
2
LIGNINGER
1
Forbind de ligninger og løsninger der hører sammen.
2x = – 8
–x =
3–
1x
2
4x
1
1
= 2x – 4 2
x=8+
–2
5x = 2
5
9x
3x – 1 =
+x
8
1,5
1
2
5
3x = 1,
4
3
–1
1
– 23x
–11 2 =
–4
–3x = 3
7,5
2x = 4
2
Udfyld de tomme figurer, så udsagnene bliver sande.
a
–7 12 + 8 =
Hvis figurerne i
en opgave er ens,
skal der stå det samme
tal i figurerne
· 3 = 33
e
1
2
2
11
i
15 · 15 = 225
j
1
2
k
1 :
25
b
f
10 – 11 = –1
c
–1 + 5 = 4
d
6 · 2,5 = 15
50
= 12
2 :6=
g
h 119
1
3
: 7 = 17
l
1
1
2
·
1
2
= 14
=2
= 1
1
3
Forbind ligninger med samme løsning.
4x =
3x + 8
30 +
2
4
Løs ligningerne.
a
x + 7,3 = 14,2
8
2
= 20
2x + 6 =
c
4x – 2
30
x – 18,3 = 9,87
x + 14,1 = 29
d
e
0,3x = 9
3x =
x = 30
17
ALINEA
–8 +
–8
2
x – 0,08 = 3,99
x = 4,07
f
x = 14,9
M AT E M AT R I X 7
= 30
x = 28,17
x = 6,9
b
3x = 20
3x – 2 = –
6
x = 30 –
6,5 x = 0,13
x = 0,02
LIGNINGER
1
Find den ubekendte:
a
2a + 1 = 9
a=
4
e
3z – 12 = 57
z=
23
b
2y + 2 = 12
y=
5
f
x + 56 = 34
x=
–22
c
2b – 2 = 12
b=
7
g
34 + 6n = 52
n=
3
d
12 – 4c = – 8
c=
5
h
26 + 2p = 16
p=
–5
2
Find den ubekendte:
a 2a
= 1
a=
b 2x
= 7
x=
y
6
= 13
y=
d b7
= 42
b=
c
2
14
78
294
36
e 3z
4
= 27
z=
f
5b
7
= 35
b=
g
3p
5
= 12
p=
20
h
4n
5
= 20
n=
25
49
3
Løs ligningerne:
a
9x –15 + 2x + 28 = 46
x=
b
12x – 14 + 12 = 10 + 4x
x=
c
3x + 4 + 2 = 2 – 5x + 20
x=
4
På en skålvægt er der ligevægt, når der ligger 2 kugler på den ene skål og 3 klodser på den anden.
Sæt streg under de situationer, hvor der også er ligevægt.
3
1,5
M AT E M AT R I X 7
2
d
24 + 15x – 16 – 9x = 50
x=
e
10x – 15 – 12 + 2x = 9
x=
f
5 + 5x – 10 = 10x + 10
x=
7
3
–3
g
4(2x – 3) – x = –5
x=
h
–3(2 – x) + 8x = 16
x=
i
8(x + 3) – 21 = 67
x=
1
2
8
a
6 kugler og 8 klodser
g
0 kugler og 0 klodser
b
4 kugler og 6 klodser
h
24 kugler og 48 klodser
c
1 kugle og 2 klodser
i
8 kugler og 12 klodser
d
12 kugler og 18 klodser
j
24 kugler og 36 klodser
e
38 kugler og 57 klodser
k
100 kugler og 150 klodser
f
15 kugler og 21 klodser
l
50 kugler og 75 klodser
18
ALINEA
LIGNINGER
1
Hvad er der gået galt i disse omformninger? Find ligningens rigtige løsning.
a
5x + 3 = 10x – 7
3 = 5x – 7
10 = 5x
10 = 5x ⇔ x = 105 ⇔ x = 2
5 = x
Den rigtige løsning er:
1
b 2 x
2
+ 2 = 8 + x
1
2 = 8 + 12 x ⇔ –6 = 12x ⇔
2 = 8 + 2 x
10 = 2 x
20 = x
c
2x – 4 = 7x + 1
–4 = 5x + 1
–4 = 5x + 1 ⇔ –4 – 1 = 5x + 1 – 1 ⇔
–3 = 5x
–5 = 5x ⇔ x = –5
– 35 = x
Den rigtige løsning er:
d
6x – 1 = 3x + 8
6x – 1 = 3x + 8 ⇔ 6x – 3x = 8 + 1 ⇔
9x – 1 = 8
3x = 9 ⇔ x =
9x = 9
⇔x=3
x = 1
Den rigtige løsning er:
e
0,5x + 4 = 4x – 3
4 = 3,5x – 3
7 = 3,5x
1 = x
2
Løs ligningerne:
a
5x + 6 = 7x – 2
b
4x – 12 + 6x = 5 – x
x = 0,5
c
5y + 5 = 7 + y
y = 0,5
1
Den rigtige løsning er:
–5
3
7
7 = 3,5x ⇔ x = 315
⇔x=2
Den rigtige løsning er:
2
x=4
x=1
9z + 2 – 6z = 6 + z – 4
z=0
M AT E M AT R I X 7
–12
9
3
+ 4 – x = 2 12 + x
d 12 x
e
–6 · 2 = 12x · 2 ⇔ x = –12
y=2
f
15y + 6 – 32 y = 32 + 12 y
g
35z + 12 – 15z = 10z – 18
z = –3
h
20z + 5 + 12,5z = 16 – 0,5z
z=
i
25b + 3 + 10b = 70 – b + 5
b=2
j
12 + 35 – 9x = x – 10 + 7
x=5
19
ALINEA
1
3
LIGNINGER
1
a
Find højden i en trekant, når arealet er 32 cm2, og grundlinjen er 8 cm.
h=
b
Opstil en ligning, der kan bruges til at finde højden (h) i en trekant,
når arealet er A cm2 og grundlinjen er g cm.
32 · 2
8
h=
c
h
g
A·2
g
Omform ligningen og find grundlinjen (g) i en trekant,
når arealet er 15 cm2 og højden er 12 cm.
g=
2
cm = 8 cm
A·2
h
15 · 2
12
g =
I en trekant er den korteste side x cm lang, den anden side er dobbelt
så lang og den tredje side er 9 cm længere end den korteste.
a
Opstil en ligning, der kan bruges til at beregne trekantens omkreds O.
O=
b
Beregn omkredsen af trekanten, når den korteste side er 4 cm.
cm = 2,5 cm
x + 2x + (x + 9) = 4x + 9
4 · 4 + 9 cm = 25 cm
c
Find længderne af alle sider, når omkredsen er 41 cm.
4x + 9 = 41 cm ⇔ 4x = 32 cm ⇔ x = 8 cm
Side 1:
8
cm
16
Side 2:
3
Omskriv formlen for trapezets areal.
a
h=
2 · (a A+ b)
b
(a + b) =
2·
A
h
c
a=
2·
A
h
2·
A
h
cm
h
b
–b
A=
–a
b=
4
I et parallelogram er den ene side 6 cm og den tilsvarende højde er 4 cm.
Beregn den anden højde (h2), når den anden side har længden 8 cm.
A = 6 · 4 = 24 cm2
h2 ·
cm
a
d
17
Side 3:
24
8
1
2
· h · (a + b)
h
=3
A=h·g
M AT E M AT R I X 7
20
ALINEA
LIGNINGER
Uligheder
1
Forbind ulighederne med de rigtige løsningsmængder.
4x + 3 <
x>3
3x +
10x + 5
1>
10
x<2
3x – 6 <
2x > 10
x>5
5x + 2
< 65
x<4
x<6
x>4
x
3x + 5x <
3–
16
x<–
2x >
1
3x –
> 22
x<5
2x < 8
12
9
2
Løs ulighederne:
a
x–2<6
x<8
e
12 + 5x ≥ 4x – 6
b
3y ≥ 90
y ≥ 30
f
6x – 3,6 < 2,4
c
9y < 36 + 6y
y < 12
g
2z + 16 ≤ 2 + z
d
4x – 2 < 22
x<6
h
24 – 4x + 12 > 8x + 12
d
y≤9
e
q ≥ –6
f
a≥0
x<3
x ≥ –18
x<1
z ≤ –14
x<2
Indtegn ulighederne på tallinjer:
a
x>5
b
z < –4
c
p≤–2

1 2 3 4 5 6 7 8 9 10 11 12

–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3

–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3
3
Løs ulighederne:
a
2x + 2 < 3x – 12
b 12 y
>4–y
c
x + 7 – x < 2(x + 6)
d
4(3x – 5) > 4(x + 3)
e 12 (4x
f
+ 2) ≤ 3(4x – 3)
–13 + 7x > –x + 19
M AT E M AT R I X 7
x > 14

–1 0 1 2 3 4 5 6 7 8 9 10

–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5
x < –11
g
3(2x – 1) > 3x + 4(x + 2)
h
1 + 7x + 12 (2x + 2) > 14x – 28
x<5
i
4(2x + 1) < 2(2x + 8)
x<3
x>4
j
–6(2x + 1) ≥ 11x – 29
x≤1
x≥1
k
7(3 + 5x) < 19x – 53
x < –2
x>4
l
–(7x + 5) ≥ 8(5 – 2x)
x≥5
y>
8
3
x > –2,5
21
ALINEA
LIGNINGER