Ta có \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)

\(\Leftrightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)

\(\Leftrightarrow25x^2-30xy+9y^2-16z^2=9x^2-30xy+25y^2\)

\(\Leftrightarrow16x^2=16y^2+16z^2\Leftrightarrow x^2=y^2+z^2\)

(5x - 3y + 4z) . (5x - 3y - 4z) = (3x - 5y)²

(5x - 3y)² - 16z² = (3x - 5y)²

25x² - 2.5x.3y + 9y² - 16z² = 9x² - 2.3x.5y + 25y²

16x² + 9y² - 16z² - 25y² = 0

16x² - 16y² - 16z² = 0

x² - y² - z² = 0

x² = y² + z²

Ta có:

\(x^2-y^2-z^2=0\)

\(16x^2-16y^2-16z^2=0\)

\(25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)

\(\left(5x-3y\right)^2-16z^2= \left(3x-5y\right)^2\)

\(\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)

Vì x² - y² - z² = 0 => x² - y² = z²

Biến đổi vế trái ta có:

(5x-3y+4z)(5x-37-4z)=(3x-5y)² - 16z²

=25x² - 30xy + 9y² - 16(x² - y²)

= 25x² - 30xy + 9y² - 16x² + 16y²

= 9x² - 30xy + 25y²

= (3x-5y)² (đpcm)

Ta có

\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-16z^2\)

\(=25x^2-30xy+9y^2-16z^2\left(!\right)\)

Thay \(x^2=y^2+z^2\) vào ! thì

\(25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)

\(=\left(3x-5y\right)^2\)

Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)

Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)

\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)

\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)

ta có 
(5x - 3y + 4z)(5x - 3y - 4z) = (5x - 3y)² - 16z² 
= 25x² - 30xy + 9y² - 16x² + 16y² 
= 25y² - 30xy + 9x² = (5y - 3x)² = (3x - 5y)²

lần sau suy nghĩ kĩ rồi hãy post lên nhé ^^ 
ta có 
(5x - 3y + 4z)(5x - 3y - 4z) = (5x - 3y)² - 16z² 
= 25x² - 30xy + 9y² - 16x² + 16y² 
= 25y² - 30xy + 9x² = (5y - 3x)² = (3x - 5y)²

Ta có: 

 (5x – 3y + 4z)( 5x –3y –4z)  = (5x – 3y )² –16z²=  25x² –30xy + 9y² –16 z²

Vì         x²=y² + z² nên 25x² –30xy + 9y² –16 (x² –y²)  =  (3x –5y)²