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 Eˆ D =60o edhe ADˆ E = 60o pra ΔADE barabrinjws. Nga tw dhwnat AE = 1m EB= AB-AE ⇒ EB = 3m – 1m ⇒ EB = 2m EB = CD brinje te kundwrta tw paralelogramit EBCD. Rrjedhimisht CD = 2m
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
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 1 1 MN = ( AD+ BC ) ⇒ MN = ( 10 + 20 ) 2 2 1 MN = .30 ⇒ MN = 15 cm 2 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 = 2 y 3x = 2 y ⇒ ⇒ ⇒ -3y = 2y -30 ⇒ -3y-2y =-30 ⇒ y =6 x + y = 10 − 3 x − 3 y = −30 1 ( x + y) = 5 2 3 x = 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 1ˆ = 45 0 = 2ˆ ΔMCH dybrinjwnjwshwm pra MH = MC HC2 = 2MC2 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 1 EF = ( AD + BC ) 2 AB = AG + GD AB = 5 + 7,5 ⇒ AD =12,5 cm 1 7,5 = ( 12,5 + BC ) ⇒ 15 = 12,5 +BC ⇒ 15 – 12,5 = BC ⇒ BC = 2,5 cm 2 Pwrgjigje : Ushtrimi 8 f.54 Shwnojmw x brinjwn anwsore te trapezit e cila wshtw e barabartw me vien e mesme 1 x = (B+b) ( B – bazw e madhe, b – bazw e vogwl ) 2 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 Aˆ M = B Cˆ 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 0 − 4ˆ Mga : 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 ) = 2y2 x2 +xy + 2xy + 2y2 –x2 – 3xy = 2y2 2y2 = 2y2 b) ( x4 + y4 ) ( x + y ) – ( x3 +y3 ) xy = x5 + y5 x5 + x4y + xy4 + y5 – x4y – xy4 = x5 + y5 x5 + y5 = x5 + y5 c) ( m2 +mn + n2 ) ( m2 - mn + n2 ) = m4 +m2n2 + n4 m4 – m3n + m2n2 + m3n – m2n2 + mn3 + m2n2 – mn3 + n4 = m4 +m2n2 + n4 m4 +m2n2 + n4 = m4 +m2n2 + n4 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) 6a3x – 12a2x2 + 18ax = 6ax ( a2 – 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) x2 – xy – 8x + 8y = x ( x – y ) – 8 ( x – y ) = ( x – y ) ( x – 8 ) e) x3 + x2 + x + 1 = x2 ( x + 1 ) + ( x + 1 ) = ( x + 1 ) ( x2 + 1 ) f) x3 + 3x2 + x + 3 = x2 ( x + 3 ) + ( x + 3 ) = ( x + 3 ) ( x2 + 1 ) g) 5a – 5b + xa - xb = 5 ( a – b ) + x ( a – b ) = ( a – b ) ( 5 + x ) h) a2m - amx + x2 – ax = am ( a – x ) – x ( a – x ) = ( a – x ) ( am – x ) i) 2x3 – 2x2 – x + 1 = 2x2 ( x -1 ) – ( x – 1 ) = ( x – 1 ) ( 2x2 – 1 ) Ushtrimi 7 f.59 a) b) c) d) e) f) g)
Ushtrimi 8 f.59 a) x3 + 6x2 + 9x = x ( x2 + 6x + 32 ) = x ( x + 3 )2 b) 2x2 + 4x + 2 = 2 ( x2 + 2x +12 ) = ( vazhdo ) c) 4a2 - b2 - 2a + b = [( 2a )2 – b2 ] – ( 2a – b ) = ( 2a – b ) ( 2a + b ) – ( 2a - b ) = ( vazhdo ) d) x3 - 16x = x ( x2 – 42 ) = ( vazhdo ) e) m5 – 10m3 + 25 m = m ( m4 – 10m2 + 25 ) = m [(m2)2 – 2.5.m2 + 52 ] = ( vazhdo ) f) a2 – b2 + 2a - 2b = ( a2 – b2 ) + 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 - f)A = R - { − 6} 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) x2 – 5x + 6 = 0 ⇒ D = 52 – 4.1.6 ⇒ D = 25 – 24 ⇒ D= 1 > 0 − (−5) ± 1 ⇒ x1 = 3; x 2 = 2 x1;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 ) ⇒ x + 1 = 0 ⇒ x = −1 Pra A = R - { 0;−1;1} x −1 = 0 ⇒ x = 1 Ushtrimi 4 f.62 x2 − x x ( x − 1) x = = a) 2 x − 1 ( x + 1)( x − 1) x + 1 x2 − 9 ( x + 3)( x − 3) x + 3 = = b) 2 x−3 x − 6x + 9 ( x − 3) 2 x 2 − y 2 ( x + y )( x − y ) x + y = = c) 2 x( x − y ) x x − xy a 3 − 2a 2 a 2 (a − 2) a2 = = d) (a + 2)(a − 2) a + 2 a2 − 4 25 − a 2 (5 + a)(5 − a ) = = −(5 + a) e) a −5 − (5 − a ) m 7 − m10 m 7 (1 − m 3 ) = = −m 5 f) 5 2 2 3 m −m − m (1 − m ) Ushtrimi 5 f.63 p2 − q2 ( p + q )( p − q) p + q = = a) p( p − q) p( p − q) p 4a 2 − 9b 2 (2a + 3b)(2a − 3b) 2a − 3b = = b) a (2a + 3b) a 2a 2 + 3ab 2 2 a −b (a + b)(a − b) a + b = = c) ax − bx x ( a − b) x
Matematikë 9 Edmond Lulja Neritan Babamusta
c 6 − c 4 c 4 (c 2 − 1) c 2 (c + 1)(c − 1) = 2 = = c 2 (c − 1) d) 3 2 c +1 c +c c (c + 1) e)
x2 −1 ( x + 1)( x − 1) x − 1 = = 2 x +1 x + 2x + 1 ( x + 1) 2
f)
15 x 2 − 15 y 2 15( x + y )( x − y ) 15( x + y ) = = 2 2 x− y x − 2 xy + y ( x − y) 2
x 2 − 2 x + 1 ( x − 1) 2 = = −( x − 1) = 1 − x g) 1− x − ( x − 1) Ushtrimi 7 f.63 ax 2 − 2 x 2 x 2 (a − 2) x = = Pwr x ≠ 0 dhe a ≠ 2. Pra vlera e thyesws nuk varet nga lera e a 2ax − 4 x 2 x(a − 2) 2 10ab − 15b 2 10b 2 − 15ab = 4a 2 − 6ab 4ab − 6a 2 5b(2a − 3b) 5b(2b − 3a ) = 2a (2a − 3b) 2a (2b − 3a ) 5b 5b = 2a 2a Ushtrimi 6 f.65 a) 1 + x + x2 +
x3 1 = 1− x 1− x
(1 − x)(1 + x + x 2 ) x3 1 + = 1− x 1− x 1− x 3 3 1− x + x 1 = 1− x 1− x 1 1 = 1− x 1− x 2ax + x 2 a2 b) a + x = a+x a+x (a + x)(a + x) 2ax + x 2 a2 − = a+x a+x a+x 2 2 2 a + 2ax + x − 2ax − x a2 = a+x a+x 2 2 a a = a+x a+x 4 − 2x + x 2 − 6x c) −2− x = x+2 x+2
Matematikë 9 Edmond Lulja Neritan Babamusta 4 − 2 x + x 2 (2 + x)( x + 2) − 6 x − = x+2 x+2 x+2 2 2 4 − 2 x + x − (4 + 2,2, x + x ) − 6 x = x+2 x+2 2 2 4 − 2x + x − 4 − 4x − x − 6x = x+2 x+2 − 6x − 6x = x+2 x+2 2 a5 −1 d) a – a + a – a + 1 = a +1 a +1 (a 4 − a 3 + a 2 − a + 1)(a + 1) 2 a5 −1 − = a +1 a +1 a +1 5 4 3 2 4 3 a − a + a − a + a + a − a + a2 − a +1 2 a5 −1 − = a +1 a +1 a +1 5 5 a +1− 2 a −1 = a +1 a +1 5 5 a −1 a −1 = a +1 a +1 4
3
2
Ushtrimi 5 f.67 3x 2x 3 : = a) 2x − 2 x − 1 4 3x x −1 3 ⋅ = 2( x − 1) 2 x 4 3 3 = 4 4 a2 − b2 a − b a + b b) 2 : = c −d2 c−d c+d (a + b)(a − b) c − d a + b ⋅ = (c + d )(c − d ) a − b c + d a+b a+b = c+d c+d a 2 b − 4b 3 a 2 b a + 2b c) = 2 2 3 3ab a − 2ab
