6.linearne jednacine i nejednacine zadaci
TRANSCRIPT
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1
2
2
4
42
2
1
4
−=−
=
=−
=−
x
x
x
x
Linearne jednačine i nejednačine
A) 93 =x B) x76 −= V) 2
1
4
1=− x
3
3
9
=
=
x
x
7
6
7
6
−=
−=
x
x
A) xxx =+− 364 B) 2/02
1⋅=
+y
1
6
6
66
634
=
=
=
=−+
x
x
x
xxx
1) aa3
12
2
1=− Pazi:
B
CAC
B
A ⋅=⋅
12
1223
2123
6/31
2
2
32
2
=
=−
=−
⋅=−
=−
a
aa
aa
aa
aa
1
01
−=
=+
y
y
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2
2) 10/12,05,0 ⋅=− x
2
12
2
5
2
5
52
5102
1025
−=−=⇒−
=
=−
−=−
=−
xx
x
x
x
3
4
43
44
44
31444
3)1(1)22(24
4
34)1(
4
14)22(
2
144
4/4
3)1(
4
1)22(
2
11
4
9
3
1)1(
100
25)22(
10
51
4
12
3
1)1(25,0)22(5,01
−=
=−
=+−
+−=−
+−=−−
+−=+−
⋅+−⋅=+⋅⋅−
⋅+−⋅=+⋅−
⋅+−⋅=+⋅−
⋅+−⋅=+⋅−
x
x
xx
xx
xx
xx
xx
xx
xx
xx
To jest 3
11−=x
5
3
15
153
14122
12142
12)1(1)2(2
4/1
3
4
1
2
2
=
=
=
−+=+
=++−
=++−
⋅=+
+−
p
p
p
pp
pp
pp
pp
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3
60105512253615
52512366010515
)15(5)13(1260)7(15
12
1560
5
136060
4
760
60/12
15
5
13
1
1
4
7
−+−−=+−
−−−=+−
+−−=+−
+⋅−
−⋅=+
−⋅
⋅+
−−
=+−
xxx
xxx
xxx
xxx
xxx
7
4
28
284
=
=
=
x
x
x
�����������������
22 )32()53(6)2)(2(5 +−−=−+− xxxx
Razlika kvadrata kvadrat binoma kvadrat binoma
( ) ( )
269251243095
912425309265
)9124()25309(6205
3322)2(5532)3(6)4(5
222
222
222
22222
+−+=+++−
−−−+−=−
++−+−=−−
+⋅⋅+−+⋅⋅−=−−
xxxxx
xxxxx
xxxxx
xxxxx
1
42
42
4242
=
=
=
x
x
x
9162516402416
1616402592416
16)4(45253342)4(
16)45()34(
22
22
2222
22
−−=−+−
−+−=+−
−+⋅⋅−=+⋅⋅−
−−=−
xxxx
xxxx
xxxx
xx
0
16
0
016
=
=
=
x
x
x
055326
0)515()326(
0)5()1()3()2(
22
22
=−++−+−−
=+−−−+−−
=−⋅−−−⋅−
xxxxxx
xxxxxx
xxxx
1
56
−=
+−=
x
x
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4
2
2 2
2 2
( 1)( 1) ( 1) 5 4
1 ( 2 1) 5 4
1 2 1 5 4
x x x x
x x x x
x x x x
− + − + = −
− − + + = −
− − − − = −
11542 ++=+− xx
2
7
72
=
=
x
x
A) 0)1)(13( =+− xx
Da se podsetimo: 00 =⇔=⋅ ABA ili 0=B
Dakle:
3
1
13
13
=
=
−
x
x
x
ili 1
1
−=
+
x
x
B) 0)3)(2)(1(4 =−−+ xxx
1
1
−=
+
x
x ili
2
2
=
−
x
x ili
3
3
=
−
x
x
A) =−−− )32(5)32( xxx
Zajednički (ide ispred zagrade) )5)(32( −−= xx
B)
0)5)(32(
0)32(5)32(
=−−
=−−−
xx
xxx
032 =−x ili 05 =−x 32 =x 5=x
2
3=x
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5
Neka je X traženi broj
→=−+
3
2
14
11
x
xMnožimo unakrsno
3(11 ) 2(14 )
33 3 28 2
3 2 28 33
5 5
5
5
1
x x
x x
x x
x
x
x
+ = −
+ = −
+ = −
= −
−=
= −
Neka je X- traženi broj
30/1532
⋅+=++ xxxx
30
303031
303061015
=
=−
+=++
x
xx
xxxx�� ��� ��
Uzastopne prirodne brojeve možemo obeležiti sa ,n ,1+n ,2+n 3+n
Dakle:
4
1008
10084
610144
101464
1014321
=
=
−=
=+
=++++++
n
n
n
n
nnnn
⇒= 252n Traženi brojevi su: 252,253,254,255
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6
5
2
2
1
Neka je X-broj učenika. Ako su u odeljenju 7
3 učenika devojčice, to nam govori da su
7
4učenika dečaci.
28
)1(/28
2843
4283
7/7
44
7
3
=
−⋅−=−
−=−
=+
⋅=+
x
x
xx
xx
xx
Neka je X-broj godina koji prodje:
Majka 27 27+x
Sin 3 3+x
sad Posle x godina
Dakle:
3
4
12
124
15275
27515
27)3(5
=
=
=
−=−
+=+
+=+
x
x
x
xx
xx
xx
Da proverimo: Kroz 3 godine majka ima 27+3=30 godina a sin 3+3=6 godina. Tad je majka 5 puta starija od sina jer je
3056 =⋅
350 m puta puta Neka je X dužina celog puta. Razmišljamo: Na koji deo puta se odnosi 350m?
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7
10
1
10
91
10
9
10
54
2
1
5
2=−⇒=
+=+
Dakle, 350m se odnosi na 10
1puta
mx
x
x
3500
10350
35010
1
=
⋅=
=⋅
Primena pitagorine teoreme:
( )
cmh
h
h
h
h
hhh
hh
bha
a
a
a
a
a
aaa
aa
a
8
4
32
324
4364
3644
4436
22
12
2
22
222
2
22
2
=
=
=
−=
=+
++=+
+=+
=+
cmc
c
c
c
c
ccc
cc
cba
c
cb
cma
25
2
50
50
502
1492
1249
)1(7
?
1
7
22
222
222
___________
=−−
=
−=
−=−
−−=−
=+−+
=−+
=+
=
−=
=
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8
3
4
43
<
<
x
x
3
2
6
62
26
<−−
<
−>−
>
x
x
x
x
2
1
12
4/4
1
2
1
4
1
2
1
−≥
≤−
⋅−
≤−
x
x
x
x
Stari obim je baO 22 += Novi obim je 111 22 baO +=
1 1 12 2 62
2( 5) 2( 2) 62
2 10 2 4 62
4 62 10 4
4 48
12 12 3 15
15
O a b
b b
b b
b
b
b cm a
a cm
= + =
+ + + =
+ + + =
= − −
=
= ⇒ = + =
=
Nejednačine
A) B) V)
−∞ ∞
4( , )
3x∈ −∞ ( ,3)x∈ −∞ 1
[ )2
x∈ − ∞−∞ ∞ −∞ ∞4
3
1
2−3
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9
4<a
{ }3,2,1,0,1,2,344 −−−∈⇒<<− aa
0 1 2 3 4-1-2-3-4
A)
5 2 2 1
5 2 1 2
3 3
3
3
1
x x
x x
x
x
x
− < +
− < +
<
<
<
( ,1)x∈ −∞1
B)
2
21
4256
2546
10/2,05,04,06,0
>
>
+−>−
−>−
⋅−>−
x
x
xx
xx
xx
(2, )x∈ ∞2
5
55
+−≤y
y
→⋅+
−≤ 5/5
5
11
5 yypazi na - ispred zagrade
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10
2
17
2
15
4
30
304
2555
5525
)5(525
≥
+≥
−−
≥
−≤−
−−≤+−
−−≤
+−≤
y
y
y
y
yy
yy
yy
1[7 , )2
y∈ ∞−∞ ∞172
A)
0,8 0,8 ( 5) 0, 2 / 10
8 8 ( 5) 2
8 8 40 2
8 2 8 40
8 50
50 25 16
8 4 4
x
x
x
x
x
x x x
− ⋅ − ≥ − ⋅
− ⋅ − ≥ −
− + ≥ −
− ≥ − − −
− ≥ −
−≤ ⇒ ≤ + ⇒ ≤−
1( ,6 ]
4x∈ −∞
164
B)
1
)1/(1
01
3/03
1
>
−−<−
<−
⋅<−
p
p
p
p
(1, )p∈ ∞1
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11
1
2
2
22
352
523
<
<
<
−
+−<
x
x
x
x
x
6
2
12
122
5172
172
−>−
>
<−
−<−
<−
x
x
x
x
x
12
628864
288664
)2(1)1(8)1(64
24/24
2
3
1
4
1
6
>
+−>−−+
−++>+−
−++>−−
⋅−
++
>−
−
x
xxxx
xxxx
xxxx
xxxx
(12, )x∈ ∞12
6 3
1 3 / 123 2 4
12 4( 6) 6 36 3(3 )
12 4 24 6 36 9 3
4 6 3 36 9 12 24
5 57
57
5
2115
x x x
x x x
x x x
x x x
x
x
x
− ++ − ≤ + ⋅
+ − − ≤ + +
+ − − ≤ + +
− − ≤ + − +
− ≤
≥ −
≥ −
2[ 11 , )
5x∈ − ∞
2115
−
17523 <+−< x Ovde moramo rešavati dve nejednačine: I Spojimo rešenja )1,6(16 −∈⇒<<− xx
( 6,1)x∈ −-6 1
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12
Najpre rešimo datu nejednačinu:
2 2
2 2
2 2
( 1) ( 1) 10
( 2 1) ( 2 1) 10
2 1 2 1 10
4 10
4 10
3 10
10
3
133
x x x
x x x x x
x x x x x
x x
x x
x
x
x
− − + < − −
− + − + + < − −
− + − − − < − −
− < − −
− + < −
− < −
−>−
> +
Najmanji prirodni broj je 4=x
1(3 , )3
x∈ ∞133
2 3 4 5
Šta znači da je neki izraz pozitivan? Pa to znači da mora biti veći od nule. Dakle:
2
2 2
2 2
(3 1) ( 2) 3 ( 1) 0
(3 6 2) 3 ( 2 1) 0
3 6 2 3 6 3 0
6 6 2 3
11 5
5
11
x x x
x x x x x
x x x x x
x x x
x
x
+ ⋅ − − ⋅ + >
− + − − ⋅ + + >
− + − − − − >
− + − > +
− >
< −
5
( , )11
x∈ −∞ −5
11−
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13
Kada kažemo da neki izraz ‘nije veći’ to znači da je manji ili je jednak )(≤
Dakle:
1
5
5
55
165164
164165
16)4(4)5(1
8/22
4
8
5
−≤
−≤
−≤
+−−≤+
−≤+−+
−≤−−+⋅
⋅−≤−
−+
x
x
x
xx
xx
xx
xx
( , 1]x∈ −∞ −-1