Math, asked by lovelygirls82, 9 days ago

helpppppppppppp meeeeeeeee please ​

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Answered by Anonymous
37

3.

(i) (3a + 4b + 5c)²

using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

= (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)

= 9a² + 16b² + 10c² + 24ab + 40bc + 30ac

(ii) (4a -2b - 3c)

{4a + (-2b) + (-3c)}

= (4a)² + (-2b)² + (-3c)² + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)

= 16a² + 4b² + 9c² - 16ab + 12bc - 24ac

(iii) (2x - y + z)²

= (2x + (-y) + z)²

using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

= (2x)² + (-y)² + (z)² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)

= 2x² + y² + z² - 4xy -2yz + 4zx

#NAWABZAADI

Answered by XxNawabzaadaxX
22

(i) (3a + 4b + 5c)²

using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

= (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) +2(3a)(5c)

= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

(ii) (4a - 2b - 3c)²

[4a + (-2b) + (-3c)]²

= (4a)² + (-2b)² + (-3c)² + 2(4a)(-2b) + 2(-2b)(-3c) + 2(4a)(-3c)

= 16a² + 4b² + 9c² - 16ab + 12bc - 24ac

(iii) (2x - y + z)²

[2x + (-y) + z)²

= (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)

= 4x² + y² + z² - 4xy - 2yz + 4xz

(iv) (2x + 3y + z)²

= (2x)² + (3y)² + (z)² + 2(2x)(3y) + 2(3y)(z) + 2(2x)(z)

= 4x² + 9y² + z² + 12xy + 6yz + 4xz

#NAWABZAADA

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