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Chapter 9 व्यंजकों का गुणनखण्ड Ex दक्षता 9 Solutions

Question - 1 : - निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए :
(i) 4a – 12
(ii) ac – be + c + cb
(iii) 36P +45 P
(iv) y_2 – 8ay
(v) 7a +7b
(vi) 3a_3x – 45a_2x – 18a

Answer - 1 : -

(i) 4a – 12
हल :
4a – 12
= 4 (a-3)

(ii) ac – be + c + cb
हल :
ac – bc + c + cb
= (a – b + c + b)
= c (a + c)

(iii) 36P +45 P
हल :
36P + 45P3
(4 + 5P
2)

(iv) y2 –8ay
हल :
y2 – 8ay = 9P
= y(y – 8a)

(v) 7a +7b
हल :
7a + 7b
= 7(a + b)

(vi) 3a3x –45a2x – 18a
हल :
3a3x – 45a2x –18a
= 3a(a – 15ax – 6)

 

Question - 2 : -
निम्नलिखित को गुणनखंड की सहायता से सरल कीजिए



Answer - 2 : -



Question - 3 : - निमनलिखिते व्यंजनों के गुणनखंड कीजिए
(i) xy (z
_2 + a_2)- x_2za – y_2za
(ii). p_3 + p + q – 1 – p_2 – pq
(iii). 2ab_2 – aby + 2cby – cy_2
_(vi) x_2 +y_3 + xy (y + 1)

Answer - 3 : -

(i) xy (z2 +a2)- x2za – y2za

हल :
= xy (z2 + a2) – x2za –y2za
= xyz
2 + xya2 –x2za – y2za
= xyz
2 – x2za +xya– y2za
= xy(yz – xa) + ya(xa – yz)
(yz – xa) (xy – ya)

(ii). p3 +p + q – 1 – p2 – pq
हल :
p3 + p + q – 1 – p2 –pq
= p
3 – pq + p – p2 +q – 1
= p(p
2 – q + 1) – 1(p2 –q + 1)
= (p
2 – q + 1) (p – 1)

(iii). 2ab2 –aby + 2cby – cy2
हल :
2ab2 – aby + 2cby – cy2
= 2ab
2 – aby + 2cby – cy2
= ab(2b – y) + cy (2b – y)
= (2b – y) (ab + cy)

(vi) x2 +y3 +xy (y + 1)
हल :
x2 + y3 +xy(y + 1)
= x
2+ y3 + xy + xy
= x
+ xy + xy2 +y3
= x(x +y) + y
2(x +y)
= (x + y) (x + y
2)

 

Question - 4 : - निम्नलिखित के मान गुणनखंड की सहायता से ज्ञात कीजिए
(i) 23 x 72 +77 x 72
(ii) 56 × 25 – 25 × 39 – 25 × 17
(iii) 27 × 47 + 55 × 8 + 27 × 53 + 45 × 8

Answer - 4 : -

(i) 23 x72 +77 x 72
हल :
23 × 72 + 77 × 72
= 72(23 + 77)
= 72 × 100 = 7200

(ii) 56 × 25 – 25 × 39 – 25 × 17
हल :
56 × 25 – 25 × 39 – 25 × 17
25(56 – 39 – 17)
= 25 × (56 – 56)
= 25 × 0 = 0

(iii) 27 × 47 + 55 × 8 + 27 × 53 + 45 × 8
हल :
27 × 47 + 55 × 8 + 27 × 53 + 45 × 8
= 27 × 47 + 27 × 53 + 55 × 8 + 45 × 8
= 27 × (47 + 53) + 8(55 + 45)
= 27 × 100 + 8 × 100
= 2700 + 800 = 3500

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