Bu nima va nega kerak
Sport ichimliklarini ishlab chiqaruvchi kompaniya mahsulotini ikki xil turdagi idishda chiqaradi: katta kub shaklidagi idish (yon tomoni a a a sm) va kichik kub shaklidagi idish (yon tomoni b b b sm). Sanoat kasrlarini hisoblashda ikki kubning ayirmasini tez ko'paytuvchilarga ajratish zarur bo'ladi: masalan, a 3 − b 3 a^3 - b^3 a 3 − b 3 ifodasini algebrik manipulyatsiya bilan soddalashtirish. Aynan shu masala kublar formulalari orqali bir qadamda yechiladi.
Kublar yig'indisi va ayirmasi formulalari matematikaning ko'p sohasida ishlatiladi: kasrlarni soddalashtirish, tenglamalar yechish va DTMdagi murakkabroq algebraik ifodalar bilan ishlash. Bu formulalarni o'zlashtirgan o'quvchi ko'p bosqichli hisoblashlarni bir necha soniya ichida bajaradi.
Asosiy g'oya
Oldin bir savol: ( a + b ) 3 (a+b)^3 ( a + b ) 3 va a 3 + b 3 a^3+b^3 a 3 + b 3 bir xilmi? Keling, tekshirib ko'raylik.
( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
Ko'rinib turibdiki, a 3 + b 3 a^3+b^3 a 3 + b 3 bilan aynan bir xil emas — orada 3 a 2 b + 3 a b 2 3a^2b + 3ab^2 3 a 2 b + 3 a b 2 qo'shimcha hadlar bor. Demak, a 3 + b 3 ≠ ( a + b ) 3 a^3+b^3 \neq (a+b)^3 a 3 + b 3 = ( a + b ) 3 . Unda a 3 + b 3 a^3+b^3 a 3 + b 3 qanday ko'paytuvchilarga ajraladi?
Ko'paytmani ochib tekshiramiz. Faraz qilaylik, a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3 = (a+b)(a^2-ab+b^2) a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) deb ta'kidlaylik. O'ng tomonni ochamiz:
( a + b ) ( a 2 − a b + b 2 ) = a ⋅ a 2 − a ⋅ a b + a ⋅ b 2 + b ⋅ a 2 − b ⋅ a b + b ⋅ b 2 (a+b)(a^2-ab+b^2)
= a\cdot a^2 - a\cdot ab + a\cdot b^2 + b\cdot a^2 - b\cdot ab + b\cdot b^2 ( a + b ) ( a 2 − ab + b 2 ) = a ⋅ a 2 − a ⋅ ab + a ⋅ b 2 + b ⋅ a 2 − b ⋅ ab + b ⋅ b 2
= a 3 − a 2 b + a b 2 + a 2 b − a b 2 + b 3 = a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a 3 − a 2 b + a b 2 + a 2 b − a b 2 + b 3
= a 3 + ( − a 2 b + a 2 b ) ⏟ 0 + ( a b 2 − a b 2 ) ⏟ 0 + b 3 = a 3 + b 3 ✓ = a^3 + \underbrace{(-a^2b + a^2b)}_{0} + \underbrace{(ab^2 - ab^2)}_{0} + b^3 = a^3 + b^3 \quad \checkmark = a 3 + 0 ( − a 2 b + a 2 b ) + 0 ( a b 2 − a b 2 ) + b 3 = a 3 + b 3 ✓
Xuddi shunday yo'l bilan kublar ayirmasini ham tekshirish mumkin:
( a − b ) ( a 2 + a b + b 2 ) = a 3 + a 2 b + a b 2 − a 2 b − a b 2 − b 3 = a 3 − b 3 ✓ (a-b)(a^2+ab+b^2)
= a^3+a^2b+ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3 \quad \checkmark ( a − b ) ( a 2 + ab + b 2 ) = a 3 + a 2 b + a b 2 − a 2 b − a b 2 − b 3 = a 3 − b 3 ✓
Shunday qilib, ikkita muhim formula:
a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) \boxed{a^3+b^3 = (a+b)(a^2-ab+b^2)} a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 )
a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) \boxed{a^3-b^3 = (a-b)(a^2+ab+b^2)} a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 )
Eslab qolish uchun qoida:
Birinchi omil: ikki hadning yig'indisi yoki ayirmasi — a ± b a \pm b a ± b .
Ikkinchi omil — uchhadli: a 2 a^2 a 2 , keyin o'rta had (yig'indida minus , ayirmada musbat a b ab ab ), keyin b 2 b^2 b 2 .
Yechilgan misollar
1-misol. x 3 + 8 x^3 + 8 x 3 + 8 ni ajratish
1-qadam. Har bir hadni kub sifatida yozing:
x 3 + 8 = x 3 + 2 3 x^3 + 8 = x^3 + 2^3 x 3 + 8 = x 3 + 2 3
2-qadam. Bu kublar yig'indisi: a = x a = x a = x , b = 2 b = 2 b = 2 .
3-qadam. Formulani qo'llang:
( x + 2 ) ( x 2 − x ⋅ 2 + 2 2 ) (x+2)\bigl(x^2 - x\cdot 2 + 2^2\bigr) ( x + 2 ) ( x 2 − x ⋅ 2 + 2 2 )
4-qadam. Hisoblang:
= ( x + 2 ) ( x 2 − 2 x + 4 ) = (x+2)(x^2 - 2x + 4) = ( x + 2 ) ( x 2 − 2 x + 4 )
Tekshirish: ( x + 2 ) ( x 2 − 2 x + 4 ) = x 3 − 2 x 2 + 4 x + 2 x 2 − 4 x + 8 = x 3 + 8 (x+2)(x^2-2x+4) = x^3-2x^2+4x+2x^2-4x+8 = x^3+8 ( x + 2 ) ( x 2 − 2 x + 4 ) = x 3 − 2 x 2 + 4 x + 2 x 2 − 4 x + 8 = x 3 + 8 ✓
2-misol. 27 − a 3 27 - a^3 27 − a 3 ni ajratish
1-qadam. Kublarni aniqlang:
27 − a 3 = 3 3 − a 3 27 - a^3 = 3^3 - a^3 27 − a 3 = 3 3 − a 3
2-qadam. Bu kublar ayirmasi: a 0 = 3 a_0 = 3 a 0 = 3 (kattaroq), b 0 = a b_0 = a b 0 = a (o'zgaruvchi).
3-qadam. Formula: ( a 0 − b 0 ) ( a 0 2 + a 0 b 0 + b 0 2 ) (a_0 - b_0)(a_0^2 + a_0 b_0 + b_0^2) ( a 0 − b 0 ) ( a 0 2 + a 0 b 0 + b 0 2 ) :
( 3 − a ) ( 9 + 3 a + a 2 ) (3 - a)(9 + 3a + a^2) ( 3 − a ) ( 9 + 3 a + a 2 )
4-qadam. Javob:
= ( 3 − a ) ( a 2 + 3 a + 9 ) = (3-a)(a^2 + 3a + 9) = ( 3 − a ) ( a 2 + 3 a + 9 )
3-misol. 8 x 3 + 125 8x^3 + 125 8 x 3 + 125 ni ajratish
1-qadam. 8 x 3 = ( 2 x ) 3 8x^3 = (2x)^3 8 x 3 = ( 2 x ) 3 va 125 = 5 3 125 = 5^3 125 = 5 3 ekanini aniqlang.
2-qadam. a = 2 x a = 2x a = 2 x , b = 5 b = 5 b = 5 bilan kublar yig'indisi formulasi:
3-qadam. ( a + b ) = ( 2 x + 5 ) (a+b) = (2x+5) ( a + b ) = ( 2 x + 5 ) va a 2 − a b + b 2 = ( 2 x ) 2 − ( 2 x ) ( 5 ) + 5 2 a^2 - ab + b^2 = (2x)^2 - (2x)(5) + 5^2 a 2 − ab + b 2 = ( 2 x ) 2 − ( 2 x ) ( 5 ) + 5 2 :
4-qadam. Hisoblang:
8 x 3 + 125 = ( 2 x + 5 ) ( 4 x 2 − 10 x + 25 ) 8x^3 + 125 = (2x+5)(4x^2 - 10x + 25) 8 x 3 + 125 = ( 2 x + 5 ) ( 4 x 2 − 10 x + 25 )
4-misol. 64 a 3 − 27 b 3 64a^3 - 27b^3 64 a 3 − 27 b 3 ni ajratish
1-qadam. 64 a 3 = ( 4 a ) 3 64a^3 = (4a)^3 64 a 3 = ( 4 a ) 3 va 27 b 3 = ( 3 b ) 3 27b^3 = (3b)^3 27 b 3 = ( 3 b ) 3 .
2-qadam. Kublar ayirmasi: p = 4 a p = 4a p = 4 a , q = 3 b q = 3b q = 3 b .
3-qadam.
( 4 a − 3 b ) ( ( 4 a ) 2 + ( 4 a ) ( 3 b ) + ( 3 b ) 2 ) (4a - 3b)\bigl((4a)^2 + (4a)(3b) + (3b)^2\bigr) ( 4 a − 3 b ) ( ( 4 a ) 2 + ( 4 a ) ( 3 b ) + ( 3 b ) 2 )
4-qadam.
= ( 4 a − 3 b ) ( 16 a 2 + 12 a b + 9 b 2 ) = (4a-3b)(16a^2 + 12ab + 9b^2) = ( 4 a − 3 b ) ( 16 a 2 + 12 ab + 9 b 2 )
5-misol. a 3 + b 3 a 2 − a b + b 2 \dfrac{a^3+b^3}{a^2-ab+b^2} a 2 − ab + b 2 a 3 + b 3 ifodasini soddalashtiring
1-qadam. Suratni ajrating: a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3 = (a+b)(a^2-ab+b^2) a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) .
2-qadam. Kasrga joylashtiring:
( a + b ) ( a 2 − a b + b 2 ) a 2 − a b + b 2 \frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2} a 2 − ab + b 2 ( a + b ) ( a 2 − ab + b 2 )
3-qadam. a 2 − a b + b 2 ≠ 0 a^2-ab+b^2 \neq 0 a 2 − ab + b 2 = 0 bo'lsa, qisqartiring:
= a + b = a + b = a + b
Mashq
a 3 + 27 a^3 + 27 a 3 + 27 ni ko'paytuvchilarga ajrating.
b 3 − 64 b^3 - 64 b 3 − 64 ni ko'paytuvchilarga ajrating.
8 + x 3 8 + x^3 8 + x 3 ni ko'paytuvchilarga ajrating.
125 a 3 − 1 125a^3 - 1 125 a 3 − 1 ni ko'paytuvchilarga ajrating.
27 x 3 + 8 y 3 27x^3 + 8y^3 27 x 3 + 8 y 3 ni ko'paytuvchilarga ajrating.
a 3 − 8 b 3 a^3 - 8b^3 a 3 − 8 b 3 ni ko'paytuvchilarga ajrating.
a 3 + 1 a 2 − a + 1 \dfrac{a^3+1}{a^2-a+1} a 2 − a + 1 a 3 + 1 ni soddalashtiring.
27 − x 3 3 − x \dfrac{27-x^3}{3-x} 3 − x 27 − x 3 ni soddalashtiring.
x 6 − 1 x^6 - 1 x 6 − 1 ni ko'paytuvchilarga ajrating. (Maslahat: x 6 = ( x 2 ) 3 x^6 = (x^2)^3 x 6 = ( x 2 ) 3 )
( a + 1 ) 3 − ( a − 1 ) 3 (a+1)^3 - (a-1)^3 ( a + 1 ) 3 − ( a − 1 ) 3 ni hisoblang va soddalashtiring.
Javoblar va yechim yo'llari
a 3 + 27 = a 3 + 3 3 = ( a + 3 ) ( a 2 − 3 a + 9 ) a^3+27 = a^3+3^3 = (a+3)(a^2-3a+9) a 3 + 27 = a 3 + 3 3 = ( a + 3 ) ( a 2 − 3 a + 9 ) .
b 3 − 64 = b 3 − 4 3 = ( b − 4 ) ( b 2 + 4 b + 16 ) b^3-64 = b^3-4^3 = (b-4)(b^2+4b+16) b 3 − 64 = b 3 − 4 3 = ( b − 4 ) ( b 2 + 4 b + 16 ) .
8 + x 3 = 2 3 + x 3 = ( 2 + x ) ( 4 − 2 x + x 2 ) 8+x^3 = 2^3+x^3 = (2+x)(4-2x+x^2) 8 + x 3 = 2 3 + x 3 = ( 2 + x ) ( 4 − 2 x + x 2 ) .
125 a 3 − 1 = ( 5 a ) 3 − 1 3 = ( 5 a − 1 ) ( 25 a 2 + 5 a + 1 ) 125a^3-1 = (5a)^3-1^3 = (5a-1)(25a^2+5a+1) 125 a 3 − 1 = ( 5 a ) 3 − 1 3 = ( 5 a − 1 ) ( 25 a 2 + 5 a + 1 ) .
27 x 3 + 8 y 3 = ( 3 x ) 3 + ( 2 y ) 3 = ( 3 x + 2 y ) ( 9 x 2 − 6 x y + 4 y 2 ) 27x^3+8y^3 = (3x)^3+(2y)^3 = (3x+2y)(9x^2-6xy+4y^2) 27 x 3 + 8 y 3 = ( 3 x ) 3 + ( 2 y ) 3 = ( 3 x + 2 y ) ( 9 x 2 − 6 x y + 4 y 2 ) .
a 3 − 8 b 3 = a 3 − ( 2 b ) 3 = ( a − 2 b ) ( a 2 + 2 a b + 4 b 2 ) a^3-8b^3 = a^3-(2b)^3 = (a-2b)(a^2+2ab+4b^2) a 3 − 8 b 3 = a 3 − ( 2 b ) 3 = ( a − 2 b ) ( a 2 + 2 ab + 4 b 2 ) .
a 3 + 1 a 2 − a + 1 = ( a + 1 ) ( a 2 − a + 1 ) a 2 − a + 1 = a + 1 \dfrac{a^3+1}{a^2-a+1} = \dfrac{(a+1)(a^2-a+1)}{a^2-a+1} = a+1 a 2 − a + 1 a 3 + 1 = a 2 − a + 1 ( a + 1 ) ( a 2 − a + 1 ) = a + 1 .
27 − x 3 3 − x = ( 3 − x ) ( 9 + 3 x + x 2 ) 3 − x = 9 + 3 x + x 2 \dfrac{27-x^3}{3-x} = \dfrac{(3-x)(9+3x+x^2)}{3-x} = 9+3x+x^2 3 − x 27 − x 3 = 3 − x ( 3 − x ) ( 9 + 3 x + x 2 ) = 9 + 3 x + x 2 .
x 6 − 1 = ( x 2 ) 3 − 1 3 = ( x 2 − 1 ) ( x 4 + x 2 + 1 ) = ( x − 1 ) ( x + 1 ) ( x 4 + x 2 + 1 ) x^6-1 = (x^2)^3 - 1^3 = (x^2-1)(x^4+x^2+1) = (x-1)(x+1)(x^4+x^2+1) x 6 − 1 = ( x 2 ) 3 − 1 3 = ( x 2 − 1 ) ( x 4 + x 2 + 1 ) = ( x − 1 ) ( x + 1 ) ( x 4 + x 2 + 1 ) .
Kengaytiramiz: ( a + 1 ) 3 = a 3 + 3 a 2 + 3 a + 1 (a+1)^3 = a^3+3a^2+3a+1 ( a + 1 ) 3 = a 3 + 3 a 2 + 3 a + 1 va ( a − 1 ) 3 = a 3 − 3 a 2 + 3 a − 1 (a-1)^3 = a^3-3a^2+3a-1 ( a − 1 ) 3 = a 3 − 3 a 2 + 3 a − 1 . Ayiramiz: ( a 3 + 3 a 2 + 3 a + 1 ) − ( a 3 − 3 a 2 + 3 a − 1 ) = 6 a 2 + 2 (a^3+3a^2+3a+1)-(a^3-3a^2+3a-1) = 6a^2+2 ( a 3 + 3 a 2 + 3 a + 1 ) − ( a 3 − 3 a 2 + 3 a − 1 ) = 6 a 2 + 2 .
Keng tarqalgan xatolar
1. Kublar yig'indisini kub yig'indisi bilan chalkastirish.
✗ a 3 + b 3 = ( a + b ) 3 a^3+b^3 = (a+b)^3 a 3 + b 3 = ( a + b ) 3
✓ a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3 = (a+b)(a^2-ab+b^2) a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) ; chunki ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a+b)^3 = a^3+3a^2b+3ab^2+b^3 ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
2. Kublar ayirmasida noto'g'ri formula ishlatish.
✗ a 3 − b 3 = ( a − b ) 3 a^3-b^3 = (a-b)^3 a 3 − b 3 = ( a − b ) 3
✓ a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3-b^3 = (a-b)(a^2+ab+b^2) a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 )
3. O'rta haddagi ishoraga e'tiborsizlik.
✗ a 3 + b 3 = ( a + b ) ( a 2 + a b + b 2 ) a^3+b^3 = (a+b)(a^2\mathbf{+}ab+b^2) a 3 + b 3 = ( a + b ) ( a 2 + ab + b 2 ) — musbat qo'yish
✓ a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3 = (a+b)(a^2\mathbf{-}ab+b^2) a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) — yig'indida minus !
4. Ikkinchi omilda a 2 a^2 a 2 va b 2 b^2 b 2 ni noto'g'ri hisoblash.
✗ 8 x 3 + 125 = ( 2 x + 5 ) ( 2 x 2 − 10 x + 25 ) 8x^3+125 = (2x+5)(2x^2-10x+25) 8 x 3 + 125 = ( 2 x + 5 ) ( 2 x 2 − 10 x + 25 ) — a 2 = ( 2 x ) 2 a^2=(2x)^2 a 2 = ( 2 x ) 2 ni 2 x 2 2x^2 2 x 2 deb yozish
✓ ( 2 x ) 2 = 4 x 2 (2x)^2 = 4x^2 ( 2 x ) 2 = 4 x 2 , shuning uchun ( 2 x + 5 ) ( 4 x 2 − 10 x + 25 ) (2x+5)(4x^2-10x+25) ( 2 x + 5 ) ( 4 x 2 − 10 x + 25 )
5. Kubni noto'g'ri ajratish: 8 x 3 = 8 ⋅ x 3 8x^3 = 8\cdot x^3 8 x 3 = 8 ⋅ x 3 emas, ( 2 x ) 3 (2x)^3 ( 2 x ) 3 .
✗ 8 x 3 + 1 = ( 8 x + 1 ) ( … ) 8x^3+1=(8x+1)(\ldots) 8 x 3 + 1 = ( 8 x + 1 ) ( … ) — asosni noto'g'ri tanlash
✓ 8 x 3 = ( 2 x ) 3 8x^3 = (2x)^3 8 x 3 = ( 2 x ) 3 , demak a = 2 x a=2x a = 2 x ; to'g'ri: ( 2 x + 1 ) ( 4 x 2 − 2 x + 1 ) (2x+1)(4x^2-2x+1) ( 2 x + 1 ) ( 4 x 2 − 2 x + 1 )
DTM uslubidagi test
Quyidagi test savollari platformada interaktiv shaklda ishlaydi: variantni tanlang, to'g'ri javob va batafsil yechimi ko'rsatiladi.