If you're seeing this message, it means we're having trouble loading external resources on our website.

Agar veb-filtrlardan foydalanayotgan boʻlsangiz *.kastatic.org va *.kasandbox.org domenlariga ruxsat berilganligini tekshirib koʻring.

Asosiy kontent

Algebra I

Course: Algebra I > Unit 15

Lesson 9: Koʻpaytuvchilarga ajratish: kvadrat uchhad
Matematika
Algebra I
Chiziqli funksiyalar koʻpaytmasiga yoyish
Koʻpaytuvchilarga ajratish: kvadrat uchhad
© 2023 Khan AcademyFoydalanish shartlariMaxfiylik siyosatiCookie Notice

Koʻpaytuvchilarga ajratish: kvadrat uchhad

Google sinfxona
“Toʻla kvadrat” shaklidagi kvadrat ifodalarni koʻpaytuvchilarga ajratishni oʻrganing. Misol uchun, x²+6x+9 ifoda (x+3)² shaklida koʻpaytuvchilarga ajraladi.
Koʻphadlarni koʻpaytuvchilarga ajratish ikki yoki undan ortiq koʻphadlarning yigʻindisini yozishni taqozo qiladi. Bu koʻphadlar koʻpaytmasining teskari holatini koʻrsatadi.
Bu yerda biz kvadrat uchhadni maxsus formulalar orqali birhad shakliga olib kelishni oʻrganamiz. Bu yigʻindi va ayirmaning kvadrati usulini koʻrsatadi, demak, siz ishga kirishishdan oldin buni bilishingiz kerak.

Kirish: Qisqa koʻpaytirish formulalari bilan koʻpaytuvchilarga ajratish

Kvadratga koʻtarish uchun biz quyidagi formulalarni ishlatishimiz mumkin.
  • left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared, equals, start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared
  • left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis, squared, equals, start color #11accd, a, end color #11accd, squared, minus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared
Eʼtibor bering, formuladagi a va b istalgan algebraik ifoda boʻlishi mumkin. Misol uchun, biz left parenthesis, x, plus, 5, right parenthesis, squared ni koʻpaytuvchilarga ajratmoqchimiz. Bunda start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd va start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 5, end color #1fab54 boʻlib, quyidagi natija hosil boʻladi:
(x+5)2=x2+2(x)(5)+(5)2=x2+10x+25\begin{aligned}(\blueD x+\greenD 5)^2&=\blueD x^2+2(\blueD x)(\greenD5)+(\greenD 5)^2\\\\ &=x^2+10x+25\end{aligned}
Siz bu formulani left parenthesis, x, plus, 5, right parenthesis, squared ni koʻpaytuvchilarga ajratib tekshirishingiz mumkin.
Bu kengaytmani ortiga qaytarish formulaga solishdir. Agar biz ifodani ortiga qaytarsak, koʻphadlarni koʻpaytuvchilarga ajratish formulasini hosil qilamiz: a, squared, plus minus, 2, a, b, plus, b, squared.
  • start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
  • start color #11accd, a, end color #11accd, squared, minus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
Biz birinchi formulani mana bu x, squared, plus, 10, x, plus, 25 koʻphad uchun ishlatsak boʻladi. Bizga start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd va start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 5, end color #1fab54 berilgan.
x2+10x+25=x2+2(x)(5)+(5)2=(x+5)2\begin{aligned}x^2+10x+25&=\blueD x^2+2(\blueD x)(\greenD5)+(\greenD 5)^2\\\\ &=(\blueD x+\greenD 5)^2\end{aligned}
Bunday koʻrinishda yozish qisqa koʻpaytirish formulalari deyiladi. Bu koʻrinishdagi koʻphadlarni birhadning kvadrati shaklida yozish mumkin!
Endi bir nechta qisqa koʻpaytirish formulalari qoʻllangan misollarni koʻrib chiqamiz.

1-misol: x, squared, plus, 8, x, plus, 16 ni koʻpaytuvchilarga ajratish

Eʼtibor bering, birinchi va soʻnggi hadlar kvadrat shaklda: x, squared, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared va 16, equals, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared. Oʻrtadagi had esa birinchi va soʻnggi hadlar koʻpaytmasining ikkilangan shaklida: 2, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, equals, 8, x.
Bundan kelib chiqadiki, koʻphad qisqa koʻpaytirish formulasiga tushadi va biz uni quyidagi formula orqali ifodalashimiz mumkin.
start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd va start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 4, end color #1fab54 berilgan. Biz bu koʻphadni quyidagi tarzda koʻpaytuvchilarga ajratishimiz mumkin:
x2+8x+16=(x)2+2(x)(4)+(4)2=(x+4)2\begin{aligned}x^2+8x+16&=(\blueD x)^2+2(\blueD x)(\greenD 4)+(\greenD4)^2\\ \\ &=(\blueD{x}+\greenD{4})^2\end{aligned}
Biz misolni left parenthesis, x, plus, 4, right parenthesis, squared ni ochib chiqib tekshirishimiz mumkin:
(x+4)2=(x)2+2(x)(4)+(4)2=x2+8x+16\begin{aligned}(x+4)^2&=(x)^2+2(x)(4)+(4)^2\\ \\ &=x^2+8x+16 \end{aligned}

Mavzu boʻyicha bilimingizni tekshiring

1) x, squared, plus, 6, x, plus, 9 ni koʻpaytuvchilarga ajrating.
Bitta javobni tanlang:

2) x, squared, minus, 6, x, plus, 9 ni koʻpaytuvchilarga ajrating.
Bitta javobni tanlang:

3) x, squared, plus, 14, x, plus, 49 ni koʻpaytuvchilarga ajrating.

2-misol: 4, x, squared, plus, 12, x, plus, 9 ni koʻpaytuvchilarga ajratish

Qisqa koʻpaytirish formulalarida birinchi hadning koeffitsiyenti 1 ga teng boʻlishi shart emas.
Misol uchun, 4, x, squared, plus, 12, x, plus, 9 da birinchi va soʻnggi hadlar kvadrat shaklda: 4, x, squared, equals, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, squared va 9, equals, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, squared. Oʻrtadagi had esa birinchi va soʻnggi hadlar koʻpaytmasining ikkilangan shaklida: 2, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, equals, 12, x.
Chunki, 4, x, squared, plus, 12, x, plus, 9 ifoda ham qisqa koʻpaytirish formulalariga tushadi. Biz yana quyidagi formulani ishlatishimiz mumkin.
start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
Bu holda, start color #11accd, a, end color #11accd, equals, start color #11accd, 2, x, end color #11accd va start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 3, end color #1fab54. Koʻphad quyidagi shaklga keladi:
4x2+12x+9=(2x)2+2(2x)(3)+(3)2=(2x+3)2\begin{aligned}4x^2+12x+9&=(\blueD {2x})^2+2(\blueD {2x})(\greenD 3)+(\greenD3)^2\\ \\ &=(\blueD{2x}+\greenD{3})^2\end{aligned}
Biz misolni left parenthesis, 2, x, plus, 3, right parenthesis, squared ifodani koʻpaytuvchilarga ajratib tekshirishimiz mumkin.

Tushunganingizni tekshiring.

4) 9, x, squared, plus, 30, x, plus, 25 ni koʻpaytuvchilarga ajrating.
Bitta javobni tanlang:

5) 4, x, squared, minus, 20, x, plus, 25 ni koʻpaytuvchilarga ajrating.

Murakkab masalalar

6*) x, start superscript, 4, end superscript, plus, 2, x, squared, plus, 1 ni koʻpaytuvchilarga ajrating.

7*) 9, x, squared, plus, 24, x, y, plus, 16, y, squared ni koʻpaytuvchilarga ajrating.

Muhokamaga qoʻshilmoqchimisiz?

Kirish
Hozircha izohlar yoʻq.
Ingliz tilini tushunasizmi? Khan Academyʼning inglizcha saytida boʻlayotgan muhokamalarni koʻrish uchun shu yerga bosing.