Cho $x^2 – y^2 – z^2 =0$ Chứng minh $(5x-3y+4z)(5x-3y-4z)=(3x-5y)^2$

Đáp án:

Ta có : 

`(5x – 3y + 4z)(5x – 3y – 4z) – (3x – 5y)^2`

` = (5x – 3y)^2 – (4z)^2 – (3x – 5y)^2`

` = (5x – 3y)^2 – 16z^2 – (3x – 5y)^2`

` = [(5x – 3y)^2 – (3x – 5y)^2] – 16z^2`

` = (5x – 3y – 3x + 5y)(5x – 3y + 3x – 5y) – 16z^2`

` = (2x + 2y)(8x – 8y) – 16z^2`

` = 2(x + y).8(x – y) – 16z^2`

` = 16(x + y)(x – y) – 16z^2`

` = 16(x^2 – y^2) – 16z^2`

` = 16(x^2 – y^2 – z^2)`

`= 16.0`

`= 0 `

`=> (5x – 3y + 4z)(5x – 3y – 4z) – (3x – 5y)^2 = 0`

`=> (5x – 3y + 4z)(5x – 3y – 4z) =  (3x – 5y)^2` (đpcm)

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