8类NCERT解决方案–第14章分解–练习14.1

问题1：找出给定条件的共同因素。
(i)12倍，36
(ii)2年，22年
(iii)14pq，28p ² q ²
(ⅳ)2X，3X ^2,4
(ⅴ)6abc，体24ab ^{2，12A 2b中}
(ⅵ)16倍^{3，-4x 2，32}倍
(vii)10pq，20qr，30rp
^(ⅷ)3×2 Y ^3，10×3 Y ^2，6×2 Y ² Z

解决方案：

(i)12倍，36

Factors of 12x and 36 are
⇒ 12x = 2 × 2 × 2 × 3 × x
⇒ 36 = 2 × 2 × 3 × 3
So, common factors are
⇒ 2 × 2 × 3 × 3 = 12

(ii)2年，22年

Factors of 2y, 22xy
⇒ 2y = 2 × y
⇒ 22xy = 2 × 11 × x × y
So, common factors are
⇒ 2 × y = 2y

(iii)14pq，28p ² q ²

Factors of 14pq, 28p²q²
⇒ 14pq = 2 × 7 × p × q
⇒ 28p²q² = 2 × 2 × 7 × p × p × q × q
So, common factors are
⇒ 2 × 7 × p × q = 14pq

(ⅳ)2X，3X ^2,4

Factors of 2x, 3x², 4
⇒ 2x = 2 × x
⇒ 3x² = 3 × x × x
⇒ 4 = 2 × 2
So, common factor is 1 (∵ 1 is a factor of every number)

(ⅴ)6abc，体24ab ^{2，12A 2b中}

Factors of 6abc, 24ab², 12a²b
⇒ 6abc = 2 × 3 × a × b × c
⇒ 24ab² = 2 × 2 × 2 × 3 × a × b × b
⇒ 12a²b = 2 × 2 × 3 × a × a × b
So, common factors are
⇒ 2 × 3 × a × b = 6ab

(ⅵ)16倍^{3，-4x 2，32}倍

Factors of 16x³, -4x², 32x
⇒ 16x³ = 2 × 2 × 2 × 2 × x × x × x
⇒ -4x² = -1 × 2 × 2 × x × x
⇒ 32x = 2 × 2 × 2 × 2 × 2
So, common factors are
⇒ 2 × 2 × x = 4x

(vii)10pq，20qr，30rp

Factors of 10pq, 20qr, 30rp
⇒ 10pq = 2 × 5 × p × q + 
⇒ 20qr = 2 × 2 × 5 × q × r
⇒ 30rp = 2 × 3 × 5 × r × p
So, common factors are
⇒ 2 × 5 = 10

^(ⅷ)3×2 Y ^3，10×3 Y ^2，6×2 Y ² Z

Factors of 3x²y³, 10x³y², 6x²y²z
⇒ 3x²y³ = 3 × x × x × y × y × y
⇒ 10x³y² = 2 × 5 × x × x × x × y × y
⇒ 6x²y²z = 2 × 3 × x × x × y × y
So, common factors are
⇒ x × x × y × y = x²y²

问题2：分解以下表达式。
(i)7倍-42
(ii)6分-12分
(iii)7a ² + 14a
(iv) − 16z + 20z ³
(v) ²⁰公升2 m + 30安
(vi)5x ² y – 15xy ²
(vii)10a ² − 15b ² + 20c ²
(viii) − 4a ² + 4ab − 4ca
(ix)x ² yz + xy ² z + xyz ²
(x)的斧² Y + BXY ² + cxyz

解决方案：

(i)7倍-42

⇒ 7x = 7 × x
⇒ 42 = 2× 3 × 7
So, common factor is 7
Therefore, 7x − 42 = 7(x − 6)

(ii)6分-12分

⇒ 6p = 2 × 3 × p
⇒ 12q = 2 × 2 × 3 × q
So, common factors are 2 × 3
Therefore, 6p − 12q = 2 × 3[p − (2 × q)]
⇒ 6(p − 2q)

(iii)7a ² + 14a

⇒ 7a² = 7 × a × a
⇒ 14a = 2 × 7 × a
So, common factors are 7 × a
Therefore, 7a² + 14a = 7 × a(a + 2)
⇒ 7a(a + 2)

(iv)-16z + 20z ³

⇒ 16z = 2 × 2 × 2 × 2 × z
⇒ 20z² = 2 × 2 × 5 × z × z × z
So, common factors are 2 × 2 × z
Therefore, −16z + 20z³ = −(2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)
⇒ 2 × 2 × z[−(2 × 2) + (5 × z × z)
⇒ 4z(−4 + 5z²)

(v) ²⁰公升2 m + 30安

⇒ 20l²m = 2 × 2 × 5 × l × l × m
⇒ 30alm = 2 × 3 × 5 × a × l × m
So, common factors are 2 × 5 × l × m
Therefore, 20l²m + 30alm = 2 × 5 × l × m[(2 × l) + (3 × a)]
⇒ 10lm(2l + 3a)

(vi)5x ² y-15xy ²

⇒ 5x²y = 5×x×x×y
⇒ 15xy² = 3×5×x×y×y
So, common factors are 5×x×y
Therefore, 5x²y − 15xy² = 5×x×y[(x) − (3×y)]
⇒ 5xy(x − 3y)

(vii)10a ² − 15b ² + 20c ²

⇒ 10a² = 2×5×a×a
⇒ 15b² = 3×5×b×b
⇒ 20c² = 2×2×5×c×c
So, common factor is 5
Therefore, 10a² − 15b² +20c² = 5[(2×a×a) − (3×b×b) + (2×2×c×c)]
⇒ 5(2a² − 3b² + 4c²)

(viii)−4a ² + 4ab − 4ca

⇒ 4a² = 2×2×a×a
⇒ 4ab = 2×2×a×b
⇒ 4ca = 2×2×c×a
So, common factors are 2×2×a = 4a
Therefore, −4a² + 4ab − 4ca = 4a(−a + b − c)

(ix)x ² yz + xy ² z + xyz ²

⇒ x²yz = x×x×y×z
⇒ xy²z = x×y×y×z
⇒ xyz² = x×y×z×z
So, common factors are x×y×z = xyz
Therefore, x²yz + xy²z + xyz² = xyz(x +y + z)

(x)的斧² Y + BXY ² + cxyz

⇒ ax²y = a×x×x×y
⇒ bxy² = b×x×y×y
⇒ cxyz = c×x×y×z
So, common factors are x×y = xy
Therefore, ax²y + bxy² + cxyz = xy(ax +by +cz)

问题3：分解。
(i)x ² + xy + 8x + 8y
(ii)15xy-6x + 5y – 2
(iii)ax + bx-ay-
(iv)15pq + 15 + 9q + 25p
(v)z − 7 + 7xy − xyz

解决方案：

(i)x ² + xy + 8x + 8y

⇒ x×x + x×y + 8×x + 8×y
Assembling the terms,
⇒ x(x + y) + 8(x + y)
Therefore, the factors are
⇒ (x + y)(x + 8)

(ii)15xy-6x + 5y-2

⇒ 3×5×x×y − 2×3×x + 5×y − 2
Assembling the terms
⇒ 3x(5y − 2) + 1(5y − 2)
Therefore, the factors are
⇒ (5y − 2)(3x + 1)

(iii)ax + bx-ay-

⇒ a×x + b×x − a×y − b×y
Assembling the terms
⇒ x(a + b) − y(a + b)
Therefore, the factors are
⇒ (a + b)(x − y)

(iv)15pq + 15 + 9q + 25p

⇒ 3×5×p×q + 3×5 + 3×3×q + 5×5×p
Assembling the terms
⇒ 3q(5p + 3) + 5(5p + 3)
Therefore, the factors are
⇒ (5p + 3)(3q + 5)

(v)z − 7 + 7xy − xyz

⇒ z − 7 + 7×x×y − x×y×z
Assembling the terms
⇒ z(1 − xy) − 7(1 − xy)
Therefore, the factors are
⇒ (1 − xy)(z − 7)