Matematika

8-sinf · Algebra · 30 daqiqa

Kasrlarni qo'shish va ayirish

Avval shuni biling:Kasrni qisqartirish
Ratsional kasrlar: mavzular

Bu nima va nega kerak

Ikkita quvur orqali hovuzga suv quyilmoqda. Birinchi quvur hovuzni yolg'iz o'zi aa soatda, ikkinchi quvur esa bb soatda to'ldiradi. Birgalikda ular 1 soatda hovuzning qancha qismini to'ldiradi?

Birinchi quvur 1 soatda hovuzning 1a\dfrac{1}{a} qismini, ikkinchi quvur esa 1b\dfrac{1}{b} qismini to'ldiradi. Birgalikda:

1a+1b\frac{1}{a} + \frac{1}{b}

Bu ifodani bitta kasrga keltirish uchun oddiy kasrlardagidek umumiy maxraj topamiz:

1a+1b=b+aab\frac{1}{a} + \frac{1}{b} = \frac{b + a}{ab}

Fizikada parallel ulangan qarshiliklar (1R=1R1+1R2\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}), tezlik va mehnat masalalari hamda DTM imtihonlaridagi algebraik ifodalar aynan shu amallarga tayanadi.

Asosiy tushunchalar va qoidalar

1. Bir xil maxrajli kasrlar

Bir xil maxrajli kasrlarni qo'shish yoki ayirishda suratlari qo'shiladi (ayiriladi), maxraj esa o'zgarishsiz qoladi:

AC+BC=A+BC,AC−BC=A−BC\frac{A}{C} + \frac{B}{C} = \frac{A + B}{C}, \qquad \frac{A}{C} - \frac{B}{C} = \frac{A - B}{C}

[!WARNING] Ayirish amalini bajarganda ikkinchi kasrning butun surati ayriladi: AC−BC=A−(B)C\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A - (B)}{C}. Qavs ochilganda BB dagi barcha hadlarning ishorasi o'zgaradi!

2. Har xil maxrajli kasrlar

Har xil maxrajli kasrlarni qo'shish va ayirish ketma-ketligi:

  1. Maxrajlarni ko'paytuvchilarga ajrating: umumiy ko'paytuvchini chiqarish yoki formulalar orqali.
  2. Eng kichik umumiy maxrajni (EKUM) toping: barcha ko'paytuvchilarning eng yuqori darajalarini ko'paytiring.
  3. Qo'shimcha ko'paytuvchini toping: umumiy maxrajni har bir kasrning o'z maxrajiga bo'ling.
  4. Suratlarni ko'paytirib, yig'ing: suratlarni o'z qo'shimcha ko'paytuvchilariga ko'paytirib, umumiy maxraj ustida yozing.
  5. Ixchamlang va qisqartiring: suratdagi o'xshash hadlarni birlashtiring, imkon bo'lsa kasrni qisqartiring.

3. Ishorani boshqarish

Agar maxrajlar qarama-qarshi bo'lsa (a−ba - b va b−ab - a), b−a=−(a−b)b - a = -(a - b) qoidasidan foydalanib kasr oldidagi ishorani almashtiramiz:

Mb−a=−Ma−b\frac{M}{b - a} = -\frac{M}{a - b}

Yechilgan misollar

1-misol. Bir xil maxrajli kasrlar va qavs

3x−1x−4−2x+3x−4\frac{3x - 1}{x - 4} - \frac{2x + 3}{x - 4}

  • Maxrajlar bir xil. Suratlar ayirmasini qavs bilan yozamiz: (3x−1)−(2x+3)x−4=3x−1−2x−3x−4=x−4x−4=1(x≠4)\frac{(3x - 1) - (2x + 3)}{x - 4} = \frac{3x - 1 - 2x - 3}{x - 4} = \frac{x - 4}{x - 4} = 1 \quad (x \neq 4)

2-misol. Birhadli maxrajlar

56a2−29ab+12b2\frac{5}{6a^2} - \frac{2}{9ab} + \frac{1}{2b^2}

  • Koeffitsiyentlar 6, 9, 2 uchun EKUK — 18. Harflar: a2a^2 va b2b^2. Umumiy maxraj: 18a2b218a^2b^2.
  • Qo'shimcha ko'paytuvchilar: mos ravishda 3b23b^2, 2ab2ab va 9a29a^2.
  • Birlashtiramiz: 5⋅3b2−2⋅2ab+1⋅9a218a2b2=9a2−4ab+15b218a2b2\frac{5 \cdot 3b^2 - 2 \cdot 2ab + 1 \cdot 9a^2}{18a^2b^2} = \frac{9a^2 - 4ab + 15b^2}{18a^2b^2}

3-misol. Kvadratlar ayirmasi qatnashgan holat

2x−3−12x2−9\frac{2}{x - 3} - \frac{12}{x^2 - 9}

  • x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3). Umumiy maxraj: (x−3)(x+3)(x - 3)(x + 3).
  • 1-kasrga qo'shimcha ko'paytuvchi: (x+3)(x + 3).
  • Yozamiz: 2(x+3)−12(x−3)(x+3)=2x+6−12(x−3)(x+3)=2(x−3)(x−3)(x+3)=2x+3\frac{2(x + 3) - 12}{(x - 3)(x + 3)} = \frac{2x + 6 - 12}{(x - 3)(x + 3)} = \frac{2(x - 3)}{(x - 3)(x + 3)} = \frac{2}{x + 3}

4-misol. Ishorani almashtirish

y2y−5+255−y\frac{y^2}{y - 5} + \frac{25}{5 - y}

  • 5−y=−(y−5)5 - y = -(y - 5) bo'lgani uchun, kasr oldidagi ishorani minusga almashtiramiz: y2y−5−25y−5=y2−25y−5=(y−5)(y+5)y−5=y+5(y≠5)\frac{y^2}{y - 5} - \frac{25}{y - 5} = \frac{y^2 - 25}{y - 5} = \frac{(y - 5)(y + 5)}{y - 5} = y + 5 \quad (y \neq 5)

5-misol. Uchta kasr yig'indisi

1a−1−2aa2−1+1a+1\frac{1}{a - 1} - \frac{2a}{a^2 - 1} + \frac{1}{a + 1}

  • Umumiy maxraj: (a−1)(a+1)=a2−1(a - 1)(a + 1) = a^2 - 1.
  • Suratlarni yig'amiz: 1(a+1)−2a+1(a−1)(a−1)(a+1)=a+1−2a+a−1a2−1=0a2−1=0(a≠±1)\frac{1(a + 1) - 2a + 1(a - 1)}{(a - 1)(a + 1)} = \frac{a + 1 - 2a + a - 1}{a^2 - 1} = \frac{0}{a^2 - 1} = 0 \quad (a \neq \pm 1)

Mashq

Amallarni bajaring va natijani qisqarmaydigan kasrga keltiring:

  1. 5xx+2+10x+2\dfrac{5x}{x + 2} + \dfrac{10}{x + 2}
  2. a2a−3−9a−3\dfrac{a^2}{a - 3} - \dfrac{9}{a - 3}
  3. 12x+34x\dfrac{1}{2x} + \dfrac{3}{4x}
  4. xx−1−1x\dfrac{x}{x - 1} - \dfrac{1}{x}
  5. 3a+2+2a−2\dfrac{3}{a + 2} + \dfrac{2}{a - 2}
  6. xx−4+44−x\dfrac{x}{x - 4} + \dfrac{4}{4 - x}
  7. a+1a−1−a−1a+1\dfrac{a + 1}{a - 1} - \dfrac{a - 1}{a + 1}
  8. 2xx2−4−1x−2\dfrac{2x}{x^2 - 4} - \dfrac{1}{x - 2}
  9. 1x2−x−1x2−1\dfrac{1}{x^2 - x} - \dfrac{1}{x^2 - 1}
  10. m−3m2−4+12−m\dfrac{m - 3}{m^2 - 4} + \dfrac{1}{2 - m}
Javoblar va yechim yo'llari
  1. 5x+10x+2=5(x+2)x+2=5\dfrac{5x + 10}{x + 2} = \dfrac{5(x + 2)}{x + 2} = 5.
  2. a2−9a−3=(a−3)(a+3)a−3=a+3\dfrac{a^2 - 9}{a - 3} = \dfrac{(a - 3)(a + 3)}{a - 3} = a + 3.
  3. Umumiy maxraj 4x4x: 2+34x=54x\dfrac{2 + 3}{4x} = \dfrac{5}{4x}.
  4. Umumiy maxraj x(x−1)x(x - 1): x2−(x−1)x(x−1)=x2−x+1x(x−1)\dfrac{x^2 - (x - 1)}{x(x - 1)} = \dfrac{x^2 - x + 1}{x(x - 1)}.
  5. Umumiy maxraj a2−4a^2 - 4: 3(a−2)+2(a+2)a2−4=5a−2a2−4\dfrac{3(a - 2) + 2(a + 2)}{a^2 - 4} = \dfrac{5a - 2}{a^2 - 4}.
  6. xx−4−4x−4=x−4x−4=1\dfrac{x}{x - 4} - \dfrac{4}{x - 4} = \dfrac{x - 4}{x - 4} = 1.
  7. (a+1)2−(a−1)2a2−1=4aa2−1\dfrac{(a + 1)^2 - (a - 1)^2}{a^2 - 1} = \dfrac{4a}{a^2 - 1}.
  8. 2x−(x+2)(x−2)(x+2)=x−2(x−2)(x+2)=1x+2\dfrac{2x - (x + 2)}{(x - 2)(x + 2)} = \dfrac{x - 2}{(x - 2)(x + 2)} = \dfrac{1}{x + 2}.
  9. 1x(x−1)−1(x−1)(x+1)=(x+1)−xx(x−1)(x+1)=1x(x2−1)\dfrac{1}{x(x - 1)} - \dfrac{1}{(x - 1)(x + 1)} = \dfrac{(x + 1) - x}{x(x - 1)(x + 1)} = \dfrac{1}{x(x^2 - 1)}.
  10. m−3(m−2)(m+2)−1m−2=m−3−(m+2)(m−2)(m+2)=−5m2−4\dfrac{m - 3}{(m - 2)(m + 2)} - \dfrac{1}{m - 2} = \dfrac{m - 3 - (m + 2)}{(m - 2)(m + 2)} = -\dfrac{5}{m^2 - 4}.

Keng tarqalgan xatolar

1. Ayirishda suratdagi hadlarning ishorasini o'zgartirmaslik. ✗ 5xx−1−2x−3x−1=5x−2x−3x−1\dfrac{5x}{x-1} - \dfrac{2x - 3}{x-1} = \dfrac{5x - 2x - 3}{x-1} ✓ Butun surat qavsga olinadi: 5x−(2x−3)x−1=5x−2x+3x−1=3x+3x−1\dfrac{5x - (2x - 3)}{x-1} = \dfrac{5x - 2x + 3}{x-1} = \dfrac{3x + 3}{x-1}

2. Maxrajlarni bir-biriga shunchaki qo'shib yuborish. ✗ ab+cd=a+cb+d\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d} ✓ Faqat umumiy maxrajga keltirib qo'shiladi: ad+bcbd\dfrac{ad + bc}{bd}.

3. Qo'shimcha ko'paytuvchini faqat bir hadga ko'paytirish. ✗ x+12+x3\dfrac{x+1}{2} + \dfrac{x}{3} da birinchi kasrni 3 ga ko'paytirganda 3x+13x + 1 deb yozish. ✓ To'g'risi butun ifodaga ko'paytiriladi: 3(x+1)=3x+33(x + 1) = 3x + 3.

4. Qarama-qarshi maxrajda ishorani almashtirishni unutish. b−ab - a o'rniga a−ba - b yozilganda, kasr oldidagi ishora ham teskarisiga o'zgarishi shart.

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Keyingi qadam

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Test

0 / 5 javob berildi

  1. 1.2xx−3−6x−3\dfrac{2x}{x - 3} - \dfrac{6}{x - 3} ifodani soddalashtiring.

  2. 2.12a+13a\dfrac{1}{2a} + \dfrac{1}{3a} yig'indini hisoblang.

  3. 3.1x−2−4x2−4\dfrac{1}{x - 2} - \dfrac{4}{x^2 - 4} ayirmani hisoblang.

  4. 4.xx−5+55−x\dfrac{x}{x - 5} + \dfrac{5}{5 - x} ifodaning qiymatini toping.

  5. 5.3a−1−2a+1\dfrac{3}{a - 1} - \dfrac{2}{a + 1} ifodani soddalashtiring.