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प्रश्न
Add:
5x2 – 3xy + 4y2 – 9, 7y2 + 5xy – 2x2 + 13
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उत्तर
We have,
(5x2 – 3xy + 4y2 – 9) + (7y2 + 5xy – 2x2 + 13)
= 5x2 – 3xy + 4y2 – 9 + 7y2 + 5xy – 2x2 + 13
= (5x2 – 2x2) + (–3xy + 5xy) + (4y2 + 7y2) + (–9 + 13) ...[Grouping like terms]
= 3x2 + 2xy + 11y2 + 4
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