Math, asked by ayushanands, 11 months ago

4a^2+9b^2+!6c^1+1/9a^2+1/14b&^2+1/25c^2=133/30;. 1440a^2b^2c^2=​

Answers

Answered by amitnrw
0

1440a²b²c² = 1 if 4a² + 9b² + 16c²  +  1/9a²  +  1/16b²  + 1/25c²  = 133/30

Step-by-step explanation:

4a² + 9b² + 16c²  +  1/9a²  +  1/16b²  + 1/25c²  = 133/30

=> (2a)² + (1/3a)²  + (3b)² + (1/4b)²  + (4c)² + (1/5c)² = 133/30

=> (2a - 1/3a)² + 2(2a)(1/3a)  + (3b - 1/4b)² + 2(3b)(1/4b) + (4c - 1/5c)² + 2(4c)(1/5c) = 133/30

=> (2a - 1/3a)²  + 4/3  + (3b - 1/4b)² + 3/2  + (4c - 1/5c)² + 8/5 = 133/30

=>  (2a - 1/3a)²  + 40/30  + (3b - 1/4b)² + 45/30  + (4c - 1/5c)² + 48/30 = 133/30

=> (2a - 1/3a)²  + (3b - 1/4b)²  + (4c - 1/5c)² + 133/30 = 133/30

=> (2a - 1/3a)²  + (3b - 1/4b)²  + (4c - 1/5c)² = 0

=> (2a - 1/3a)² = 0  => 2a - 1/3a = 0  => 6a² = 1

(3b - 1/4b)²  = 0  => 3b - 1/4b = 0  => 12b² = 1

(4c - 1/5c)² = 0 => 4c - 1/5c = 0  => 20c² = 1

(6a²)( 12b²)(20c²) = 1 * 1 * 1

=> 1440a²b²c² = 1

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4a^2+9b^2+16c^2+1/9a^2+1/16b^2+25c^2=133/30. ;1440a^2b^2c^2=​

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