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

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

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

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

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

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

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

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

Cách 1:x²-y²-z²=0

=>x²=y²+z²

(5x-3y+4z)(5x-3y-4z)

=(5x-3y)²-16z²

=25x²-30xy+9y²-16z²(*)

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

Cách 2: cách này dễ hiểu hơn

x²-y²-z²=0

=>x²=y²+z²

(5x-3y+4z).(5x-3y-4z)=(3x-5y)²

<=>(5x-3y)²-16z²=(3x-5y)²

<=>(5x-3y)²-(3x-5y)²=16z²

<=>(8x-8y)(2x+2y)=16z²

<=>16(x²-y²)=16z²

<=>x²=y²+z² (đúng với gt)

Ta có: (5x-3y+4z)(5x-3y-4z)=(5x-3y)^2-16z^2=25x^2-30xy+9y^2-16(x^2-y^2)=25x^2-30xy+9y^2-16x^2+16y^2

                                                                                                            =9x^2-30xy+25y^2=(3x-5y)^2  (đpcm)
 

            Bài làm :

Ta có:

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

\(\Leftrightarrow16x^2-16y^2-16z^2=0\)

\(\Leftrightarrow25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)

\(\Leftrightarrow\left[\left(25x^2-30xy+9y^2\right)-16z^2\right]-\left(9x^2-30xy+25y^2\right)=0\)

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

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

=> Điều phải chứng minh

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ta có:

\(x^2-y^2-z^2=0\left(gt\right)\)

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

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

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

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

\(\Rightarrow\left(2x+2y\right)\left(8x-8y\right)=16z^2\)

\(\Rightarrow2\left(x+y\right).8\left(x-y\right)=16z^2\)

\(\Rightarrow16\left(x^2-y^2\right)=16z^2\)

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

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

\(\Rightarrow\) Đúng với giả thuyết ban đầu

Vậy \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\) với \(x^2-y^2-z^2=0\)