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Factorise: (a2 − 1) (b2 − 1) + 4ab - Mathematics

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प्रश्न

Factorise:

(a2 − 1) (b2 − 1) + 4ab

योग
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उत्तर

(a2 − 1) (b2 − 1) + 4ab

= a2b2 − a2 − b2 + 1 + 4ab

= a2b2 + 1 + 2ab − a2 − b2 + 2ab

= (a2b2 + 1 + 2ab) − (a2 + b2 − 2ab)

= (ab + 1)2 − (a − b)2

= [(ab + 1) − (a − b)][(ab + 1) + (a − b)]     ...[∵ a2 − b2 = (a + b)(a − b)]

= [ab + 1 − a + b][ab + 1 + a − b]

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Method of Factorisation : Difference of Two Squares
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Factorisation - Exercise 5 (C) [पृष्ठ ७३]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 5 Factorisation
Exercise 5 (C) | Q 20 | पृष्ठ ७३
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