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

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1. Đáp án:

    

   Giải thích các bước giải:

   Xét: $(5x-3y+4z)(5x-3y-4z)-(3x-5y)^2$

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

   Thay $x^2 – y^2 – z^2 = 0$
   $= 16 . 0 $ 

   $= 0$
   $⇒(5x-3y+4z)(5x-3y-4z)- (3x-5y)^2 = 0$

   $⇒(5x-3y+4z)(5x-3y-4z) = (3x – 5y)^2 $ (điều phải chứng minh)

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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)

   Giải thích các bước giải:

    

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