מתוך ויקיספר, אוסף הספרים והמדריכים החופשי
בחלק קטן מהתרגילים הבאים, ידע קודם בפירוק מספר לגורמים ראשוניים יכול לעזור.
פישוט ביטויים [ עריכה ] פשט את הביטויים הבאים:
a
5
⋅
a
9
a
12
{\displaystyle {\frac {a^{5}\cdot a^{9}}{a^{12}}}}
פתרון
a
5
⋅
a
9
a
12
=
a
5
+
9
a
12
=
a
14
a
12
=
a
14
−
12
=
a
2
{\displaystyle {\frac {a^{5}\cdot a^{9}}{a^{12}}}={\frac {a^{5+9}}{a^{12}}}={\frac {a^{14}}{a^{12}}}=a^{14-12}=a^{2}}
a
7
⋅
a
10
⋅
a
20
⋅
a
6
{\displaystyle a^{7}\cdot a^{10}\cdot a^{20}\cdot a^{6}}
פתרון
a
7
⋅
a
10
⋅
a
20
⋅
a
6
=
a
7
+
10
+
20
+
6
=
a
43
{\displaystyle a^{7}\cdot a^{10}\cdot a^{20}\cdot a^{6}=a^{7+10+20+6}=a^{43}}
(
3
7
)
10
⋅
(
3
8
)
12
(
3
5
)
33
{\displaystyle {\frac {(3^{7})^{10}\cdot (3^{8})^{12}}{(3^{5})^{33}}}}
פתרון
(
3
7
)
10
⋅
(
3
8
)
12
(
3
5
)
33
=
3
7
⋅
10
⋅
3
8
⋅
12
3
5
⋅
33
=
3
70
⋅
3
96
3
165
=
3
166
3
165
=
3
166
−
165
=
3
1
=
3
{\displaystyle {\frac {(3^{7})^{10}\cdot (3^{8})^{12}}{(3^{5})^{33}}}={\frac {3^{7\cdot 10}\cdot 3^{8\cdot 12}}{3^{5\cdot 33}}}={\frac {3^{70}\cdot 3^{96}}{3^{165}}}={\frac {3^{166}}{3^{165}}}=3^{166-165}=3^{1}=3}
16
9
2
35
{\displaystyle {\frac {16^{9}}{2^{35}}}}
פתרון
16
9
2
35
=
(
2
4
)
9
2
35
=
2
4
⋅
9
2
35
=
2
36
2
35
=
2
36
−
35
=
2
1
=
2
{\displaystyle {\frac {16^{9}}{2^{35}}}={\frac {(2^{4})^{9}}{2^{35}}}={\frac {2^{4\cdot 9}}{2^{35}}}={\frac {2^{36}}{2^{35}}}=2^{36-35}=2^{1}=2}
32
9
⋅
128
15
⋅
8
17
64
19
⋅
16
20
{\displaystyle {\frac {32^{9}\cdot 128^{15}\cdot 8^{17}}{64^{19}\cdot 16^{20}}}}
פתרון
32
9
⋅
128
15
⋅
8
17
64
19
⋅
16
20
=
(
2
5
)
9
⋅
(
2
7
)
15
⋅
(
2
3
)
17
(
2
6
)
19
⋅
(
2
4
)
20
=
2
5
⋅
9
⋅
2
7
⋅
15
⋅
2
3
⋅
17
2
6
⋅
19
⋅
2
4
⋅
20
=
2
45
⋅
2
105
⋅
2
51
2
114
⋅
2
80
=
2
45
+
105
+
51
2
114
+
80
=
2
201
2
194
=
{\displaystyle {\frac {32^{9}\cdot 128^{15}\cdot 8^{17}}{64^{19}\cdot 16^{20}}}={\frac {(2^{5})^{9}\cdot (2^{7})^{15}\cdot (2^{3})^{17}}{(2^{6})^{19}\cdot (2^{4})^{20}}}={\frac {2^{5\cdot 9}\cdot 2^{7\cdot 15}\cdot 2^{3\cdot 17}}{2^{6\cdot 19}\cdot 2^{4\cdot 20}}}={\frac {2^{45}\cdot 2^{105}\cdot 2^{51}}{2^{114}\cdot 2^{80}}}={\frac {2^{45+105+51}}{2^{114+80}}}={\frac {2^{201}}{2^{194}}}=}
2
201
−
194
=
2
7
=
128
{\displaystyle 2^{201-194}=2^{7}=128}
(
a
11
⋅
b
18
⋅
c
10
)
6
⋅
(
a
7
⋅
b
13
⋅
c
19
)
14
(
a
20
⋅
b
36
⋅
c
40
)
8
{\displaystyle {\frac {(a^{11}\cdot b^{18}\cdot c^{10})^{6}\cdot (a^{7}\cdot b^{13}\cdot c^{19})^{14}}{(a^{20}\cdot b^{36}\cdot c^{40})^{8}}}}
פתרון
(
a
11
⋅
b
18
⋅
c
10
)
6
⋅
(
a
7
⋅
b
13
⋅
c
19
)
14
(
a
20
⋅
b
36
⋅
c
40
)
8
=
(
(
a
11
)
6
⋅
(
b
18
)
6
⋅
(
c
10
)
6
)
⋅
(
(
a
7
)
14
⋅
(
b
13
)
14
⋅
(
c
19
)
14
)
(
a
20
)
8
⋅
(
b
36
)
8
⋅
(
c
40
)
8
=
.
.
.
{\displaystyle {\frac {(a^{11}\cdot b^{18}\cdot c^{10})^{6}\cdot (a^{7}\cdot b^{13}\cdot c^{19})^{14}}{(a^{20}\cdot b^{36}\cdot c^{40})^{8}}}={\frac {((a^{11})^{6}\cdot (b^{18})^{6}\cdot (c^{10})^{6})\cdot ((a^{7})^{14}\cdot (b^{13})^{14}\cdot (c^{19})^{14})}{(a^{20})^{8}\cdot (b^{36})^{8}\cdot (c^{40})^{8}}}=...}
נזכור כי כפל מקיים חוק החילוף ולכן במונה אני יכול לשים את ה- aים אחד ליד השני, את ה- bים אחד ליד השני ואת ה- cים אחד ליד השני. הכפל גם קיבוצי ולכן אני יכול לשים סוגריים מסביב לכל זוג ואז להשתמש בתכונה השנייה של חזקות שכתוב בדף הקודם.
a
66
⋅
b
108
⋅
c
60
⋅
a
98
⋅
b
182
⋅
c
266
a
160
⋅
b
288
⋅
c
320
=
a
98
+
66
⋅
b
108
+
182
⋅
c
60
+
266
a
160
⋅
b
288
⋅
c
320
=
{\displaystyle {\frac {a^{66}\cdot b^{108}\cdot c^{60}\cdot a^{98}\cdot b^{182}\cdot c^{266}}{a^{160}\cdot b^{288}\cdot c^{320}}}={\frac {a^{98+66}\cdot b^{108+182}\cdot c^{60+266}}{a^{160}\cdot b^{288}\cdot c^{320}}}=}
a
164
⋅
b
290
⋅
c
326
a
160
⋅
b
288
⋅
c
320
=
a
4
⋅
b
2
⋅
c
6
{\displaystyle {\frac {a^{164}\cdot b^{290}\cdot c^{326}}{a^{160}\cdot b^{288}\cdot c^{320}}}=a^{4}\cdot b^{2}\cdot c^{6}}
16
73
⋅
81
16
8
98
⋅
27
21
{\displaystyle {\frac {16^{73}\cdot 81^{16}}{8^{98}\cdot 27^{21}}}}
פתרון
16
73
⋅
81
16
8
98
⋅
27
21
=
(
2
4
)
73
⋅
(
3
4
)
16
(
2
3
)
98
⋅
(
3
3
)
21
=
2
292
⋅
3
64
2
294
⋅
3
63
=
2
292
−
294
⋅
3
64
−
63
=
2
−
2
⋅
3
=
1
2
2
⋅
3
=
3
4
{\displaystyle {\frac {16^{73}\cdot 81^{16}}{8^{98}\cdot 27^{21}}}={\frac {(2^{4})^{73}\cdot (3^{4})^{16}}{(2^{3})^{98}\cdot (3^{3})^{21}}}={\frac {2^{292}\cdot 3^{64}}{2^{294}\cdot 3^{63}}}=2^{292-294}\cdot 3^{64-63}=2^{-2}\cdot 3={\frac {1}{2^{2}}}\cdot 3={\frac {3}{4}}}
