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 
(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

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

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) 

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

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

\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)

\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)

\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)

\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)

\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)

câu o0o trả lời là sai

\(Đây\)\(mới\)\(là\)\(câu\)\(trả\)\(lời\)\(đúng\)

\(ta\)\(có\)\(16\left(x^2-y^2-z^2\right)=16\left(x^2-y^2\right)-16z^2=8\left(x-y\right)2\left(x-y\right)-\left(4z\right)^2=\left(8x-8y\right)\left(2x+2y\right)-\left(4z\right)^2=\left(5x-3y+3x-5y\right)\left(5x-3y-3x+5y\right)-\left(4z\right)^2\)

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

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

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

Áp dụng hằng đẳng thức ( a - b ) ( a + b ) = a² - b² ta đc:

     \(\left(5x-3y+8z\right)\left(5x-3y-8z\right)=\left(5x-3y\right)^2-\left(8z\right)^2\)

                                                                     \(=25x^2-30xy+9y^2-64z^2\)

                                                   Đề có sai ko vậy bn

mk lấy kq của bạn Kia Cerato mk giải típ

tc \(x^2=y^2+4z^2\Leftrightarrow x^2-y^2=4z^2\)

\(\Leftrightarrow25x^2-30xy+9y^2-16.4z^2\)

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

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

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

ok