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7/29/2019 23B Problem SolvingXX
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LOADING
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PROBLEM SOLVING
1. BHETY ANGGRAENI K. R 1002102040112. INTAN FIDDIYAH 100210204110
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1. 59669 x 999 + (2213) - (712) =
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Langkah 1 :
59669 x 999 =
59669
xx
= 59609331
diubah menjadi
(1000 1)59669 x
= 59669000 -
59669 x 999
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2 1 3 =2x13 2 =2
13 = +
7 36 9984
0_ _ _
04 0 25x 2
Langkah 2 :
2 13 = __04 0
9
x13 2 =
13 = +
5 36 94
25
61
x 2
2
8
Dikuadratkan
002 0
mendekati
2213
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712 = 0_06 4
4
x12 2 =
12 = +
9 34 45
91
41
x 8
Langkah 3 :
_
2
6Langkah 4 :
59669 x 999 + (2213) - (712) =
= 65166044
Dikuadratkan 08 0mendekati
712
59609331 4897369 659344+ +
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3
2735
- 2702
- 713 - 2702
2. =
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Langkah 1 :
2735 21= 2
22
= 423=
menentukan
pola 824= . . . 625= . . . 2
Berulang 4 kali
karena nilai
satuan pada 25sama dengan nilai
satuan pada 21
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Berdasarkan pola,bilangan 2 yang
dipangkatkan
memiliki 4 jenis
maka
2735735
4=183(sisa 3 )
maka
= 23= 8
2702
702
4 =175(sisa 2 )maka
= 22= 4
Langkah 2 :
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Langkah 3 :
3
2735
- 2702
+ 713+2702
= 3
8 - 4+
713+4
= 34+ 71
7
= 375
7
x
= 2175
= 725
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3. B = ; C =
, maka =
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Langkah 1 :
B =
B = 30 +dimisalkan B
B = 30 + B
B 30-B- = 0(B 6)(B + 5) VB = 6 B = -5
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Langkah 2 :
C =
C = 30 -
dimisalkan C
C = 30 - CC 30-C+ = 0
(C + 6)(C - 5)
VC = -6 C = 5
Langkah 3 :
=
=
=
6 - 530
15 - 16
130
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4. 1 + 2 + 3 + . . . + 20 =
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Langkah 1 :Menentukan pola terlebih dahulu :
1 =1 = 1 x 2( 1 + 2 )1 x 2 x 3
=
Bilangan selanjutnya
1 (dugaan)
2= 5 = 2 x 3( 2 + 3 )1 x 2 x 3
= 5
3=14= 3 x 4 ( 3 + 4 )1 x 2 x 3 = 14(pola)
1+
21+ +
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1 + 2 + 3 + . . . + 20
Langkah 2 :
Berdasarkan pola :
20 x 21 ( 20 + 21 )
1 x 2 x 3
420 ( 41)
6=
=
6
=
= 2870
4 2 0
4
1
6
1
00
8
0
0
4
0 0
02
022
1
7
X
1 +
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5. 1
x1 3
+1
x3 5
+1
x5 7
+1
x 99
+. . .
97
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1
x1 3+
1
x3 5+
1
x5 7+
1
x97 99+. . .
Langkah :
1
1
-1
3
=
. . .
1
3-1
1
3
-1
5
1
5-3
+ +1
5-
1
7
1
7-5
+
+1
97-
1
99
1
99-97
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=1
1-
1
3
1
2
+1
3
-1
5
1
2
+1
5
-1
7
1
2
+
. . . + 1
97-
1
99
1
2
=1
2
1
1-
1
3+
1
3-
1
5+
1
5-
1
7+
. . .1
97+
1
97-
1
99-
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=1
2
99
99
- 1
= 12
9899
=
49
99
=1
2
1
1-
1
99
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TERIMA KASIH
