Cho a, b, c > 0 và dãy tỉ số: 2b+c-a/a = 2c-b+a/b =2a+b-c/c Tính P = ( 3a-2b ). ( 3b-2c ). ( 3c-2a ) / ( 3a-c ). ( 3b-a ). ( 3c -b )

Ta có: `(2b+c-a)/a = (2c-b+a)/b = (2a+b-c)/c`

`=> (2b+c-a)^2 = (2c-b+a)^2 = (2a+b-c)^2`

`=> 4b^2+4c^2-4ab-4ac+a^2 = 4c^2+4a^2-4bc-4ca+b^2 = 4a^2+4b^2-4ac-4cb+c^2`

`=> 2b^2+2c^2-2ab-2ac+a^2 = 2c^2+2a^2-2bc-2ca+b^2 = 2a^2+2b^2-2ac-2cb+c^2`

`=> 2b^2-2ab+a^2 = 2c^2-2bc+b^2 = 2a^2-2ac+c^2`

`=> (2b-a)^2 = (2c-b)^2 = (2a-c)^2`

`=> (2b-a)(2c-b)(2a-c) = (2b-a)^3 = (2c-b)^3 = (2a-c)^3`

`=> (2b-a)(2c-b)(2a-c) = (2b-a)(2c-b)(2a-c)`

`=> (2b-a)(2c-b)(2a-c) = (2b-a)^3`

`=> (2b-a)(2c-b)(2a-c) = (2b-a)(2b-a)(2b-a)`

`=> (2b-a)(2c-b)(2a-c) = (3a-2b)(3b-2c)(3c-2a)`

`=> P = {(3a-2b)(3b-2c)(3c-2a)} /{ (3a-c)(3b-a)(3c-b) }= 1`

Vậy `P=1`