Chiziqli tenglamalar I. Nazariy qism
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CHIZIQLI TENGLAMALAR 010620231107
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CHIZIQLI TENGLAMALAR I. NAZARIY QISM Bir yoki bir nechta noma’lum qatnashgan tenglikka tenglama deyiladi.
2 –8x+7=0 ; x+y=5. Tenglama “=” – teng belgisidan hamda shu tenglikning chap va o‘ng qismidan tashkil topadi. Tenglamaning chap yoki o‘ng qismidagi qo‘shiluvchilar tenglamaning hadlari deyiladi. Odatda tenglamadagi noma’lumlar lotin harflari bilan belgilanadi, lekin bu qat’iy emas. Uni ixtiyoriy simvol bilan belgilab olish mumkin. Masalan, ushbu 5 ∙ ∆ − 5 = 7 ∙ ∆ − 8 ifoda ham tenglama. Tenlama o‘z tarkibida qatnashgan noma’lumlar xili va ularning darajalariga ko‘ra nomlanadi. Masalan, 1) 7x+9=2x–4 – birinchi darajali bir noma’lumli(chiziqli) tenglama; 2) 3x–y=8 – birinchi darajali ikki noma’lumli tenglama; 3) 3x
2 –5x–1=0 – ikkinchi darajali bir noma’lumli tenglama. Darsxona topshirig‘i: Quyidagi tenglamalarni nomlang: 1) x 3 –x–1=0; 2) x 2 +y–x=2; 3) 6x 2 +9x–8=6x 2 –x+3.
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Noma’lumning tenglamani to‘g‘ri tenglikka aylantiradigan qiymatiga tenglamaning ildizi(yechimi) deyiladi.
∙(–3)+1; –11= –11.
𝑥 4 −8𝑥+3 𝑥 3 −𝑥+1 + 7 = 0 tenglamaning ildizi bo‘lishini toping.
Tenglamaning barcha ildizlarini topish yoki ildizi yo‘q ekanini ko‘rsatish tenglamani yechish deyiladi.
Bir xil ildizlarga ega bo‘lgan yoki ildizi bo‘lmagan tenglamalar teng kuchli(ekvivalent) tenglamalar deyiladi. Masalan, 1) 𝑥 − 3 va 𝑥 2
𝑥+3 ;
2) 3𝑥 − 3 = 4 − 3𝑥 va 𝑥 + 1 − 5𝑥 = −4𝑥 tenglamalar teng kuchli.
bo‘ladi. ===============@grand_matematika=============== 3
Masalan, x–5=0. Tenglamaning ikkala qismiga 5 ni qo‘shamiz. x–5+5 = 5 ; x=5. 1-xossadan quyidagi natijalar kelib chiqadi: 1) tenglamaning istalgan hadini uning bir qismidan ikkinchi qismiga ishorasini qarama-qarshisiga o‘zgartirib olib o‘tish mumkin.
x= –6. 2) tenglamaning ikkala qismida aynan bir xil hadlar bo‘lsa, ularni tashlab yuborish mumkin. Masalan, x 2 +x=x 2 +17;
x=17. 2-xossa: Tenglamaning ikkala qismini noldan farqli ayni bir songa ko‘paytirilsa yoki bo‘linsa, berilgan tenglamaga teng kuchli tenglama hosil bo‘ladi. Masalan, 1) 1 3 𝑥 = 7 ;
3 ∙ 1
𝑥 = 7 ∙ 3;
𝑥 = 21. 2) 7𝑥 = 8;
7𝑥 7
8 7 ; 𝑥 =
8 7 . ===============@grand_matematika=============== 4
2-xossadan kelib chiqadigan natija: 1) tenglamada noma’lum oldidagi bo‘luvchini tenglikning boshqa qismiga ko‘paytuvchi qilib o‘tkazish mumkin. Masalan, 𝑥: 8 = 7 ; 𝑥 = 7 ∙ 8;
𝑥 = 56.
2) tenglamada noma’lum oldidagi ko‘paytuvchini tenglikning boshqa qismiga bo‘luvchi qilib o‘tkazish mumkin. Masalan, 4x=16; x=16:4; x=4. Biz kelgusida sizlar bilan tenglamaning quyidagi turlarini o‘rganishni rejalashtirganmiz: 1. Chiziqli tenglama; 2. Kvadrat tenglama; 3. Ratsional tenglama: Xususan: 1) kubik tenglama; 2) bikvadrat tenglama; 3) qaytma tenglama; 4) kasr holidagi tenglamalar. 4. Modulli tenglama; 5. Irratsional tenglama; 6. Ko‘rsatkichli tenglama; ===============@grand_matematika=============== 5
7. Logarifmik tenglama; 8. Trigonometrik tenglama; 9. Differensial tenglama; 10. Nostandart tenglamalar.
bo‘lgan tenglamalarni chiziqli tenglamalar deyiladi. Bu yerda a va b istalgan haqiqiy sonlar. a, b=const Masalan, 1) 6x=17, bunda a=6, b=17; 2) –2x = 1, bunda a= –2, b=1; 3) x=0, bunda a=1, b=0.
1. Agar { 𝑎 ≠ 0 𝑏𝜖𝑅
shart bajarilsa, tenglama bitta yechimga ega. Masalan, 4x=7 ; { 𝑎 = 4 ≠ 0 𝑏 = 7𝜖𝑅 .
Darsxona topshirig‘i: Yagona yechimga ega bo‘lgan uchta tenglama tuzing.
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2. Agar { 𝑎 = 0 𝑏 = 0
shart bajarilsa, cheksiz ko‘p yechimga ega.(Bu holda istalgan haqiqiy son tenglamaning yechimi bo‘ladi).
0 ∙x=0; { 𝑎 = 0 𝑏 = 0
. Darsxona topshirig‘i: Cheksiz ko‘p yechimga ega bo‘lgan uchta tenglama tuzing.
3. Agar { 𝑎 = 0 𝑏 ≠ 0
shart bajarilsa, yechimga ega emas. Masalan, 3x–5+2x=5x–1; 5x–5x= –1+5; 0 ∙x=4; { 𝑎 = 0; 𝑏 = 4 ≠ 0 . Darsxona topshirig‘i: Yechimga ega bo‘lmagan uchta tenglama tuzing.
Yuqorida keltirilgan tenglamaning xossalar tenglamaning barcha turi uchun amal qiladi, shu jumladan chiziqli tenglamaga ham. ===============@grand_matematika=============== 7
1-misol. 8 – 7(4x+5)=3–8x tenglamani yeching. Yechilishi: 8 – 7(4x+5)=3–8x; 8–28x–35=3–8x; –28x+8x=3–8+35; –20x=30; x= 30 −20
= − 3 2 = −𝟏, 𝟓. Darsxona topshirig‘i: 5x+9–2(1–4x)=7x+5–4(7x+3) tenglamani yeching. Javob: –0,5 2-misol. 2𝑥−3
3 − 4 = 3𝑥 + 7 − 𝑥 2
Yechilishi: 2𝑥−3
3 − 4 = 3𝑥 + 7 − 𝑥 2
6 (
2𝑥−3 3 − 4) = 6 (3𝑥 + 7 − 𝑥 2 ) ; 2(2𝑥 − 3) − 24 = 18𝑥 + 42 − 3𝑥;
4𝑥– 6 − 24 = 15𝑥 + 42;
−11𝑥 = 52;
𝑥 = −
𝟓𝟐 𝟏𝟏 . Darsxona topshirig‘i: 3𝑥 5 − 6(4 − 3𝑥) = 𝑥 + 3−𝑥
10 tenglamani yeching. Javob: 𝟐𝟒𝟑
𝟏𝟕𝟕
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Parametrli chiziqli tenglamalar Tenglmadagi o‘zgarmas sonni ifodalovchi harf yoki simvolga parametr deyiladi. Parametr ishtirok etgan chiziqli tenglamani parametrli chiziqli tenglama deyiladi. Quyida parametrli chiziqli tenglamardagi ko‘p uchraydigan asosiy misollarni ko‘rib chiqamiz:
qo‘yilgan shartga ko‘ra ish davom etadi. 4kx–9=6x+5 ; 4kx–6x=5+9; (4k–6)=14. => ax=b tenglamada yagona yechim shartini esga olamiz:
{
𝑎 ≠ 0 𝑏𝜖𝑅
=> { 4𝑘 − 6 ≠ 0 14𝜖𝑅 =>
=> 4𝑘 ≠ 6 => 𝑘 ≠ 1,5 => 𝒌𝝐(−∞; 𝟏, 𝟓) ∪ (𝟏, 𝟓 ; ∞) . Darsxona topshirig‘i: k ning qanday qiymatlarida 3x–(k+2)x–1=x+3 tenglama yagona yechimga ega? Javob: 𝒌𝝐(−∞; 𝟎) ∪ (𝟎; ∞)
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2-misol. k ning qanday qiymatida k 2 x–4x=k–2 tenglama cheksiz ko‘p yechimga ega? Yechilishi: k 2 x–4x=k–2; (k 2 –4)x=k–2. => ax=b => {𝑎 = 0 𝑏 = 0 => {𝑘
2 − 4 = 0
𝑘 − 2 = 0 =>
=> { 𝑘 = ±2
𝑘 = 2 => 𝒌 = 𝟐 . Darsxona topshirig‘i: k ning qanday qiymatida kx+7=5x–k–2 tenglama cheksiz ko‘p yechimga ega bo‘ladi? Javob: hech qanday 3-misol. k ning qanday qiymatida 2kx–5x=k+3+6x tenglama yechimga ega bo‘lmaydi? Yechilishi: 2kx–5x=k+3+6x; (2k–11)x=k+3. => ax=b => { 𝑎 = 0
𝑏 ≠ 0 => {
2𝑘 − 11 = 0 𝑘 + 3 ≠ 0 => => { 𝑘 = 5,5 𝑘 ≠ −3
=> k=5,5 . Darsxona topshirig‘i: k ning qanday qiymatida k 2 x–4x=5x+k–3 tenglama yechimga ega bo‘lmaydi? Javob: k= –3 ===============@grand_matematika=============== 10
Proporsiya Ikkita nisbatning tengligiga proporsiya deyiladi. a:b=c:d yoki 𝒂 𝒃 = 𝒄 𝒅 . Bu yerda a, d – proporsiyaning chetki hadlari, b, c – o‘rta hadlari deyiladi.
45 17 = 90 34 ; 1 7 = 5 35 ; 3𝑎 2𝑎 = 6𝑘 4𝑘 tengliklar proprsiya hisoblanadi. Darsxona topshirig‘i: Proporsiyaga uchta misol keltiring.
ko‘paytmasi o‘rta hadlari ko‘paytmasiga teng. 𝒂 𝒃 = 𝒄 𝒅 => ad=bc. Bu xossadan proporsiyaning noma’lum hadini topishda qo‘llaniladi.
3 11 = 12 𝑥 . Yechilishi: 3 11 = 12 𝑥 => 3𝑥 = 11 ∙ 12 => 𝒙 = 𝟒𝟒. Darsxona topshirig‘i: Proporsiyaning noma’lum hadini toping: 13 𝑥 = 39 24 . Javob: 8
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Mavzulashtirilgan testlar to‘plamining CHIZIQLI TENGLAMA , PROPORSIYA, PARAMETRLI CHIZIQLI TENGLAMALAR mavzusidagi testlardan namunalar. 1-test.(97-6-6) Tenglamani yeching: 6 −
𝑥−1 2 = 3−𝑥 2 + 𝑥−2 3 . A) 4,5 B) 8 C) 17 D) 11 E) 14 Yechilishi: 6 −
𝑥−1 2 = 3−𝑥 2 + 𝑥−2 3 ; 6 ∙ (6 −
𝑥−1 2 ) = 6 ( 3−𝑥 2 + 𝑥−2 3 ) ; 6 ∙ 6 − 6 ∙ 𝑥−1 2
3−𝑥 2 + 6 ∙ 𝑥−2 3 ; 36 − 3𝑥 + 3 = 9 − 3𝑥 + 2𝑥 − 4 ;
2𝑥 = 34 ;
𝑥 = 17 . Javob: 17. (C) Darsxona topshirig‘i.(97-1-6) Tenglamani yeching: 3𝑥−11
4 − 3−5𝑥 8 = 𝑥+6 2 .
A) 5 B) –4,5 C) 6,5 D) 7 E) 8 Javob: 7. (D) ===============@grand_matematika=============== 12
2-test.(97-7-3) Tenglamani yeching: 0,9(4x–2)=0,5(3x–4)+4,4 . A) 1,2 B) 2,5 C) –3 D) 2 E) 0,2 Yechilishi: 0,9(4x–2)=0,5(3x–4)+4,4 ; 3,6x–1,8=1,5x–2+4,4 ; 3,6x–1,5x= –2+4,4+1,8 ; 2,1x=4,2 ; x=2 . Javob: 2 . (D) Darsxona topshirig‘i.(96-7-3) Tenglamani yeching: 6,4(2–3x)=6(0,8x–1)+6,8 . A) 5 B) –0,5 C) 0,5 D) –2 E) 2,5 Javob: 0,5 . (C) 3-test.(01-2-59) Tenglamani yeching: 0,(3)+0,1(6) 0,(319)+1,(680) ∙ 𝑥 = 8
0,(6) .
A) 4 B) 32 C) 1 D) 1 E) 16 Yechilishi: 0,(3)+0,1(6) 0,(319)+1,(680) ∙ 𝑥 = 8
0,(6) ;
3 9 + 15 90 319 999
+1 680
999 ∙ 𝑥 = 8
6 9 ; 45 90 1 999
999 ∙ 𝑥 = (2
3 ) 2 3 ;
1 2 2 ∙ 𝑥 = 4 ; ===============@grand_matematika=============== 13
𝑥 4
𝑥 = 16 . Javob: 16. (E)
0,1(6)+0,(6) 0,(3)+1,1(6) (𝑥 + 1) = 0,3(8)𝑥 tenglamani yeching. A) 2,(6) B) –2,(6) C) 3,(6) D) –3,(6) E) –3,(3) Javob: –3,(3). (E) 4-test.(98-12-12) Tenglamani yeching: (12,5–x):5=(3,6+x):6 . A) 5 2 11 B)
5 3 11 C) 5 4 11 D)
5 1 11 E) 5 5 11
6(12,5–x)=5(3,6+x) ; 75–6x=18+5x ; 11x=57 ;
𝑥 = 57
= 5 2 11 . Javob: 𝟓 𝟐 𝟏𝟏 . (A) Darsxona topshirig‘i.(96-1-6) Proporsiyaning noma’lum hadini toping: 2 4 5 : 𝑥 = 1
2 3 : 2 6 7 . A) 1 2 B) 2 3 C) 4 4 5 D)
3 5 E) 2 1 5 Javob: 𝟒 𝟒 𝟓 . (C) ===============@grand_matematika=============== 14
5-test.(02-3-5) 𝑎 − 2𝑏 ; 4 ; 𝑎 + 3𝑏 ; 24 sonlar proporsiyaning ketma- ket hadlari bo‘lsa, 𝑎 2 −𝑏 2 2𝑎𝑏 ifodaning qiymatini toping. A) 4 3
8 3 E) 7 2
Yechilishi: 𝑎 − 2𝑏 ; 4 ; 𝑎 + 3𝑏 ; 24 𝑎−2𝑏
4 = 𝑎+3𝑏 24 ;
24(𝑎 − 2𝑏) = 4(𝑎 + 3𝑏);
24𝑎 − 48𝑏 = 4𝑎 + 12𝑏 ; 20𝑎 = 60𝑏 ;
𝑎 = 3𝑏 => 𝑎 2
2 2𝑎𝑏
= (3𝑏)
2 −𝑏 2 2∙3𝑏∙𝑏 = 8𝑏 2 6𝑏 2 = 4 3 . Javob: 𝟒 𝟑 . (A) Darsxona topshirig‘i.(var-09) Proporsiyaning dastlabki uchta hadi yig‘indisi 78 ga teng. Uning ikkinchi hadi birinchi hadining 1 2
uchinchi hadi esa 2 3 qismini tashkil qiladi. Proporsiyaning uchinchi hadini toping. A) 12 B) 18 C) 36 D) 24 Javob: 24 . (D) 6-test.(98-7-30) Ushbu10(ax–1)=2a–5x–9 tenglama a ning qanday qiymatlarida yagona yechimga ega? A) (−∞; −
1 2 ) ∪ (− 1 2 ; ∞) B) − 1 2
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C) 1 5
(−∞; − 1 2 ) E) (− 1 2 ; ∞) Yechilishi: 10(𝑎x– 1) = 2𝑎– 5x– 9 ; 10𝑎𝑥 − 10 = 2𝑎 − 5𝑥 − 9 ; ( 10𝑎 + 5)𝑥 = 2𝑎 + 1.
𝑎𝑥 = 𝑏 => 𝑎 ≠ 0 => 10𝑎 + 5 ≠ 0 =>
𝑎 ≠ −
1 2 => (−∞; − 1 2 ) ∪ (− 1 2 ; ∞) . Javob: ( −∞; −
𝟏 𝟐 ) ∪ (− 𝟏 𝟐 ; ∞) . (A) Darsxona topshirig‘i.(var-07) 2,5(ax–5,2)=2a–5x–9 tenglama a ning qanday qiymatlarida yagona yechimga ega bo‘ladi? A) − 1
B) (−∞; −
1 2 ) ∪ (− 1 2 ; ∞) C) (−∞; −2) ∪ (−2; ∞) D) 1 5
Javob: (−∞; −𝟐) ∪ (−𝟐; ∞) . (C) 7-test.(98-12-28) Tenglama a ning qanday qiymatlarida cheksiz ko‘p yechimga ega? 10(ax–1)=2a–5x–9 A) − 1
B) 2 C) 1 2 D) –2 E) 1 5 Yechilishi: 10(ax–1)=2a–5x–9 ; 10ax–10=2a–5x–9 ; (10a+5)x=2a+1 .
𝑎𝑥 = 𝑏 => { 𝑎 = 0 𝑏 = 0
=> { 10𝑎 + 5 = 0 2𝑎 + 1 = 0 => 𝑎 = −
1 2 . Javob: − 𝟏 𝟐 . (A) ===============@grand_matematika=============== 16
Darsxona topshirig‘i.(97-7-22) m ning qanday qiymatlarida m 2 x–m=x+1 tenglamaning ildizlari cheksiz ko‘p bo‘ladi? A) 1 B) 0 C) –1 D) ±1 E) ∅ Javob: –1 . (C) 8-test.(99-8-21) Tenglama a qanday qiymatlarida yechimga ega emas? 6x–a–6=(a+2)(x+2) A) 4 B) 2 C) –2 D) 6 E) –6
6x–a–6=ax+2a+2x+4 ; (4–a)x=3a+10. 𝑎𝑥 = 𝑏 => { 𝑎 = 0 𝑏 ≠ 0
=> { 4 − 𝑎 = 0 3𝑎 + 10 ≠ 0 => 𝑎 = 4 .
bo‘lmaydi? A) 1 va 7 B) 1 C) 7 D) 1 va –7 E) –7
≤ 1 D) t ≥ 2 E) 0 ===============@grand_matematika=============== 17
3x–4=2x–2t ; x=4 – 2t. => 4 – 2t >0 ; t < 2. Javob: t < 2 Darsxona topshirig‘i.(var-08) p ning qanday qiymatlarida 5x–3p=4 tenglamaning ildizi –12 dan katta bo‘ladi? A) − 56
< 𝑝 B) − 64 3 < 𝑝 C) 𝑝 < −56
3 D)
𝑝 < −64
3
− 𝟔𝟒
< 𝒑 . (B) Uyga topshiriq: Mavzulashtirilgan testlar to‘plamining CHIZIQLI TENGLAMA, PROPORSIYA, PARAMETRLI CHIZIQLI TENGLAMALAR mavzularidagi barcha testlarni yechish.
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