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

Biến đổi vế trái ta có :

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

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

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

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

\(=\left(3x-5y\right)^2\)  ( ĐPCM) 

Ta có:

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

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

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

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

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

\(x^2-y^2=4z^2\\ \Leftrightarrow64z^2=16x^2-16y^2\)

\(\left(5x-3y+8z\right)\left(5x-3y-8z\right)\\ =\left(5x-3y\right)^2-64z^2\\ =25x^2-30xy+9y^2-64z^2\\ =25x^2-16x^2+9y^2+16y^2-30xy\\ =9x^2-30xy+25y^2=\left(3x-5y\right)^2\)

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

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

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

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

Thay \(x^2=y^2+z^2\) vào ! thì

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

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