USHTRIME TE ZGJIDHURA

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Matematikë 9 Edmond Lulja Neritan Babamusta
Matematikë 9 Edmond Lulja Neritan Babamusta
Ushtrimi 4 f.52
Ndwrtojmw DE||BC
EBCD parallelogram DE = BC si brinjw tw kundwrta tw paralelogramit (1)
AD = BC nga tw dhwnat; nga (1) AD = DE (2)
Nga (2) ΔADE dybrinjwnjwshwm mga A Ê D =60 o edhe
tw dhwnat AE = 1m EB= AB-AE ⇒ EB = 3m – 1m ⇒ EB = 2m
EB = CD brinje te kundwrta tw paralelogramit EBCD. Rrjedhimisht CD = 2m
AD
ˆ E = 60 o pra ΔADE barabrinjws. Nga
Ushtrimi 5 f.52
Ndwrtojmw DE ⊥ AB dhe CF ⊥ AB
EFCD drejtkwndwsh ( DE||CF, dy drejtwza pingule me njw tw tretw; EFCD paralelogram me njw
kwnd tw drejt)
ΔADE = ΔFBC ( trekendwsha kenddrejt,AD=BC ; FC = ED) (1)
Nga (1) AE =FB dhe EF = AF – AE ⇒ EF = 30 – 6 ⇒ EF = 24 cm
EF = CD ( brinjw tw kundwrta tw drejtkwndwshit) pra CD = 24 cm
AB = 6 + 30 = 36 cm
MN = 2
1 ( AB + CD ) ⇒ MN = 2
1 ( 30 + 24 ) ⇒ MN = 2
1 .54 ⇒ MN = 27 cm
Pwrgjigje :
Ushtrimi 6 f.52
AB ⊥ d dhe BC ⊥ d pra AD||BC ( dy drejtwza pingule
me jw tw tretw janw || ndwrmjet tyra)
pra ABCD trapez.
MN ⊥ d pra MN|| AD||BC Nw bazw tw teoremws sw
Talesit MD = MC pra MN vije e mesme e trapezit
MN = 2
1 ( AD+ BC ) ⇒ MN = 2
1 ( 10 + 20 )
MN = 2
1 .30 ⇒ MN = 15 cm
Pwrgjigje :
Matematikë 9 Edmond Lulja Neritan Babamusta
Ushtrimi 7 f.52
Shwnojmw : x → baza e vogwl
y → baza e madhe
Nga tw dhwnat kemi
x 2 1
= dhe ( x + y ) = 5
y 3 2
Formojmw sistemin e ekuacioneve dhe zgjidhim
⎧ x 2
⎪ =
y 3 ⎧ 3x
= 2y
⎧ 3x
= 2y
⎨
⇒ ⎨ ⇒ ⎨
⇒ -3y = 2y -30 ⇒ -3y-2y =-30 ⇒ y =6
⎪
1
( x + y)
= 5 ⎩x
+ y = 10 ⎩−
3x
− 3y
= −30
⎩2
⎧3x
= 2.6 ⎧x
= 4
⎨ ⇒ ⎨ Pra bazat e trapezit janw 4 cm dhe 6 cm
⎩ y = 6 ⎩y
= 6
Ushtrimi 8 f.52
Shwnojmw : x → njwrwn bazw tw trapezit
x + 4 → wshtw baza tjetwr
Pwrfundimi ( 5 cm: 9cm )
Ushtrimi 5 f.54
Kushti : ABCD romb; OF ⊥ CD ; OE ⊥ BC; OH ⊥ AB; OG ⊥ AD
Pwrfundimi : OF = OE = OH = OG
ΔOCE = ΔOCF Trekwndwsha kwnddrejtw; 1ˆ = 2ˆ ; hipotenuzw tw pwrbashkwt
Pwrballw 1ˆ ndodhet OE
pwrballw 2ˆ ndodhet OF
pra OE = OF
Me tw mnjwjtin arsyetim ΔODG = ΔODF ;
ΔOAH = ΔOAG dhe ΔOBH = ΔOBE
Pwrfundimisht OF = OE = OH = OG
Ushtrimi 6 f.54
Sxhwnojmw me a brinjwn e katrorit
AC = a 2
MC = a 2 - a
0
1ˆ = 45 = 2ˆ
ΔMCH dybrinjwnjwshwm pra MH = MC
HC 2 = 2MC 2
HC = MC 2
BH = a - MC 2
BH = a – ( a 2 - a ) 2
BH = a - 2a + a 2
BH = a 2 - a = MC = MH
Matematikë 9 Edmond Lulja Neritan Babamusta
Ushtrimi 7 f.54
Ndwrtojmw EF || AD
Jemi nw kushtet e teoremws sw Talesit
AD || EF || BC caktojnw AE = EB prandaj
FC = FD kwshtu qw EF vijw e mesme e trapezit
GEFD parallelogram
EF = 7,5 cm
EF = 2
1 ( AD + BC )
AB = AG + GD
AB = 5 + 7,5 ⇒ AD =12,5 cm
7,5 = 2
1 ( 12,5 + BC ) ⇒ 15 = 12,5 +BC ⇒ 15 – 12,5 = BC ⇒ BC = 2,5 cm
Pwrgjigje :
Ushtrimi 8 f.54
Shwnojmw x brinjwn anwsore te trapezit e cila wshtw e barabartw me vien e mesme
x = 2
1 ( B + b ) ( B – bazw e madhe, b – bazw e vogwl )
2x = B + b
P = B + b + x + x ⇒ P = 2x + 2x ⇒ P = 4x, pra
24 = 4x ⇒ x = 6 cm
Pwrgjigje :
Ushtrimi 9 f.54
a) Shiko 2 f. 53
b) ABCD parallelogram MNPQ meset e brinjeve
MNPQ sipas 2 f. 53 wshtw parallelogram. A mund tw jetw drejtkwndwsh ? Diagonalet
e drejtkwndwshit janw kongruente. Gjykoni pwr MN dhe PQ, MP e PN
c) Gjykoni pwr MN e PQ, MP e PN nw se ABCD wshtw drejtkwndwsh
d) Gjykoni pwr MN e PQ, MP e PN nw se ABCD wshtw romb
e) Gjykoni pwr MN e PQ, MP e PN nw se ABCD wshtw katror
Matematikë 9 Edmond Lulja Neritan Babamusta
Pwrgjigje : a) Paralelogram; b) Paralelogram; c) Romb; d) Drejtkwndwsh; e) Katror
Ushtrimi 10 f.54
a)ΔADM dhe ΔBCD kanw : AD = BC brinje tw kundwrta tw paralelogramitl AM _ CN e dhwnw;
D Â M = BĈ N ndrrues; Rasti I i kongruencws sw trwkwndshave pra
ΔADM = ΔBCD
b) Nw mwnyrw tw ngjashme dhe ΔAMB = ΔDNC
c) Nga ΔADM = ΔBCD kemi DM = BN dhe nga ΔAMB = ΔDNC kemi BM = DN
pra brinjwt e kundwrta tw katwrkwndwshit MBND janw dy e nga dy kongruente prandaj MBND
parallelogram ( nwnyra e parw )
Nga ΔADM = ΔBCD kemi 1ˆ = 4ˆ
2ˆ = 180
0 −1ˆ
3ˆ = 180
Mga :
0 − 4ˆ
2ˆ = 3ˆ ( vazhdo vete)
Ushtrimi 11 f.54
CGEF parallelogram ( Pse ? )
CGEF drejtkwndwsh ( Pse ? )
ΔCFE kwnddrejt dybrinjwnjwsgwm ( Pse ? )
Pra CF = EF
Rrjedhimisht CGEF katror
Matematikë 9 Edmond Lulja Neritan Babamusta
Ushtrimi 5 f.57
a) ( x + 2y ) ( x + y ) – x( x + 3y ) = 2y 2
x 2 +xy + 2xy + 2y 2 –x 2 – 3xy = 2y 2
2y 2 = 2y 2
b) ( x 4 + y 4 ) ( x + y ) – ( x 3 +y 3 ) xy = x 5 + y 5
x 5 + x 4 y + xy 4 + y5 – x 4 y – xy 4 = x 5 + y 5
x 5 + y 5 = x 5 + y 5
c) ( m 2 +mn + n 2 ) ( m 2 - mn + n 2 ) = m 4 +m 2 n 2 + n 4
m 4 – m 3 n + m 2 n 2 + m 3 n – m 2 n 2 + mn 3 + m 2 n 2 – mn 3 + n 4 = m 4 +m 2 n 2 + n 4
m 4 +m 2 n 2 + n 4 = m 4 +m 2 n 2 + n 4
Ushtrimi 5 f.59
a) 15x – 30y – 45 z = 15 ( x – 2y 3z )
b) 24m -16n + 8 = 9 ( 3m -2n +1 )
c) 2am – 3cm + amx = m( 2a -3c + ax )
d) 6a 3 x – 12a 2 x 2 + 18ax = 6ax ( a 2 – 2ax + 3 )
e) 3xy3 + 6xy2 – 18 xy = 3xy ( 3y2 + 2xy – 3)
Ushtrimi 6 f.59
a) 4 ( x + y ) – m ( x + y ) = ( x + y ) ( 4 – m )
b) m ( x + 3y ) – n ( x + 3y ) = ( x + 3y ) ( m – n )
c) 7ax + 7ay – 3m ( x + y ) = 7a ( x + y ) – 3m ( x + y ) = ( x + y ) ( 7a – 3m )
d) x 2 – xy – 8x + 8y = x ( x – y ) – 8 ( x – y ) = ( x – y ) ( x – 8 )
e) x 3 + x 2 + x + 1 = x 2 ( x + 1 ) + ( x + 1 ) = ( x + 1 ) ( x 2 + 1 )
f) x 3 + 3x 2 + x + 3 = x 2 ( x + 3 ) + ( x + 3 ) = ( x + 3 ) ( x 2 + 1 )
g) 5a – 5b + xa - xb = 5 ( a – b ) + x ( a – b ) = ( a – b ) ( 5 + x )
h) a 2 m - amx + x 2 – ax = am ( a – x ) – x ( a – x ) = ( a – x ) ( am – x )
i) 2x 3 – 2x 2 – x + 1 = 2x 2 ( x -1 ) – ( x – 1 ) = ( x – 1 ) ( 2x 2 – 1 )
Ushtrimi 7 f.59
a) x 2 + 4xy + 4y 2 = ( x + 2y ) 2
b) m 4 – 2m 2 + 1 = ( m 2 -1 ) 2 = [( m + 1 ) ( m – 1 )] 2 = ( m + 1 ) 2 ( m – 1 ) 2
c) 2a 2 – 4a b + 2b 2 = 2 (a 2 – 2ab + b 2 ) = 2 ( a – b ) 2
d) 3x 3 – 12x = 3x ( x 2 – 4 ) = 3x ( x + 2 ) ( x – 2 )
e) 4x 3 – 32y 3 = 4 ( x 3 – y 3 ) = 4 (x – y ) ( x 2 + xy + y 2 )
f) 7x 5 + 7x 2 = 7x 2 ( x 3 + 1 ) = 7x 2 ( x + 1 ) ( x 2 – x + 1 )
g) 5y 3 – 5y 6 = 5y 3 ( 1 – y3 ) = 5y 3 ( 1- y ) ( 1 + x + x 2 )
Ushtrimi 8 f.59
a) x3 + 6x 2 + 9x = x ( x 2 + 6x + 3 2 ) = x ( x + 3 ) 2
b) 2x 2 + 4x + 2 = 2 ( x 2 + 2x +1 2 ) = ( vazhdo )
c) 4a 2 - b 2 - 2a + b = [( 2a ) 2 – b 2 ] – ( 2a – b ) = ( 2a – b ) ( 2a + b ) – ( 2a - b ) = ( vazhdo )
d) x 3 - 16x = x ( x2 – 4 2 ) = ( vazhdo )
e) m 5 – 10m 3 + 25 m = m ( m4 – 10m2 + 25 ) = m [(m 2 ) 2 – 2.5.m 2 + 5 2 ] = ( vazhdo )
f) a 2 – b 2 + 2a - 2b = ( a 2 – b 2 ) + 2 ( a – b ) = ( a – b ) ( a + b ) + 2 ( a – b ) = ( vazhdo )
Matematikë 9 Edmond Lulja Neritan Babamusta
Bashkwsia e vlerave tw ndryshores pwr tw cilat thyesa ka kuptim, quhet bqshkwsie e
pwrcaktimit tw thyesws.
Ushtrimi 3 f.61
Shwnojmw A bashkwsinw e pwrcaktimt tw thyesws
⎧1⎫
⎧1⎫
a) A = R – { 1 } b)A = R c)A = R d)A = R - ⎨ ⎬ e)A = R - ⎨ ⎬
⎩2⎭
⎩4⎭
⎧ 2⎫
g) A = R - ⎨ − ⎬
⎩ 7 ⎭
Ushtrimi 3 f.61
a) A = R - { 0 ;3}
b) A = R c) A = R = { 0 ;5}
d) A = R - { 1;2 }
e) x 2 – 5x + 6 = 0 ⇒ D = 5 2 – 4.1.6 ⇒ D = 25 – 24 ⇒ D= 1 > 0
− ( −5)
± 1
x 1;2 = ⇒ x1 = 3; x2
= 2
2.1
Pra A = R - { 2 ;3}
g) x3 – 4x = x( x2 – 4 ) = x ( x + 1 ) ( x – 1 )
x = 0
x ( x + 1 ) ( x – 1 ) ⇒
Ushtrimi 4 f.62
2
x − x x(
x −1)
a) =
=
2
x −1
( x + 1)( x −1)
b)
c)
d)
x + 1 = 0 ⇒ x = −1
x −1
= 0 ⇒ x = 1
x
x + 1
2
x − 9 ( x + 3)( x − 3)
=
=
2
2
x − 6x
+ 9 ( x − 3)
x
x
2 2
x − y ( x + y)(
x − y)
x + y
=
=
2
x − xy x(
x − y)
x
3
a − 2a
2
a − 4
2
+ 3
− 3
2
2
a ( a − 2) a
=
=
( a + 2)( a − 2) a + 2
2
25 − a (5 + a)(5
− a)
e) =
= −(5
+ a)
a − 5 − (5 − a)
f)
m
m
7
5
− m
− m
10
2
7 3
m (1 − m )
=
= −m
2 3
− m (1 − m )
Ushtrimi 5 f.63
2 2
p − q ( p + q)(
p − q)
p + q
a) =
=
p(
p − q)
p(
p − q)
p
b)
c)
2 2
4a − 9b
(2a
+ 3b)(2a
− 3b)
2a
− 3b
=
=
2
2a
+ 3ab
a(2a
+ 3b)
a
2 2
a − b ( a + b)(
a − b)
a + b
=
=
ax − bx x(
a − b)
x
5
Pra A = R - { 0;
− 1;1 }
f)A = R - { − 6}
Matematikë 9 Edmond Lulja Neritan Babamusta
6 4 4 2
2
c − c c ( c −1)
c ( c + 1)( c −1)
2
d) = =
= c ( c −1)
3 2 2
c + c c ( c + 1) c + 1
e)
f)
x
2
2
x −1
( x + 1)( x −1)
=
=
2
+ 2x
+ 1 ( x + 1)
x
x
−1
+ 1
2 2
15x
− 15y
15( x + y)(
x − y)
15( x + y)
=
=
2
2
2
x − 2xy
+ y ( x − y)
x − y
2
2
x − 2x
+ 1 ( x −1)
g) = = −(
x −1)
= 1−
x
1−
x − ( x −1)
Ushtrimi 7 f.63
2 2 2
ax − 2x
x ( a − 2) x
= = Pwr x ≠ 0 dhe a ≠ 2. Pra vlera e thyesws nuk varet nga lera e a
2ax
− 4x
2x(
a − 2) 2
2
10ab
−15b
2
4a
− 6ab
5b(2a
− 3b)
2a(2a
− 3b)
5b
5b
=
2a
2a
2
10b
−15ab
=
2
4ab
− 6a
5b(2b
− 3a)
=
2a(2b
− 3a)
Ushtrimi 6 f.65
3
x 1
a) 1 + x + x 2 + =
1− x 1−
x
2 3
(1 − x)(1
+ x + x ) x 1
+ =
1−
x 1−
x 1−
x
3 3
1−
x + x 1
=
1−
x 1−
x
1 1
=
1− x 1− x
2 2
2ax
+ x a
b) a + x - =
a + x a + x
2 2
( a + x)(
a + x)
2ax
+ x a
− =
a + x a + x a + x
2
2
2 2
a + 2ax
+ x − 2ax
− x a
=
a + x
a + x
2
2
a a
=
a + x a + x
2
4 − 2x
+ x − 6x
c)
− 2 − x =
x + 2
x + 2
Matematikë 9 Edmond Lulja Neritan Babamusta
2
6
2
2)
)(
(2
2
2
4
2
+
−
=
+
+
+
−
+
+
−
x
x
x
x
x
x
x
x
2
6
2
)
2,2,
(4
2
4
2
2
+
−
=
+
+
+
−
+
−
x
x
x
x
x
x
x
2
6
2
4
4
2
4
2
2
+
−
=
+
−
−
−
+
−
x
x
x
x
x
x
x
2
6
2
6
+
−
=
+
−
x
x
x
x
d) a 4 – a 3 + a 2 – a + 1 -
1
1
1
2
5
+
−
=
+ a
a
a
1
1
1
2
1
1)
1)(
(
5
2
3
4
+
−
=
+
−
+
+
+
−
+
−
a
a
a
a
a
a
a
a
a
1
1
1
2
1
1
5
2
3
4
2
3
4
5
+
−
=
+
−
+
+
−
+
−
+
+
−
+
−
a
a
a
a
a
a
a
a
a
a
a
a
a
1
1
1
2
1
5
5
+
−
=
+
−
+
a
a
a
a
1
1
1
1
5
5
+
−
=
+
−
a
a
a
a
Ushtrimi 5 f.67
a)
4
3
1
2
:
2
2
3
=
− x −
x
x
x
4
3
2
1
1)
2(
3
=
−
⋅
−
x
x
x
x
4
3
4
3 =
b)
d
c
b
a
d
c
b
a
d
c
b
a
+
+
=
−
−
−
−
:
2
2
2
2
d
c
b
a
b
a
d
c
d
c
d
c
b
a
b
a
+
+
=
−
−
⋅
−
+
−
+
)
)(
(
)
)(
(
d
c
b
a
d
c
b
a
+
+
=
+
+
c)
3
2
2
3
4
2
2
2
3
2
b
a
ab
a
b
a
ab
b
b
a +
=
−
−