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

\(A=\left(5x-3y+8x\right)\left(5x-3y-8z\right)+1\)

\(=\left(5x-3y\right)^2-64z^2+1\)

\(=\left(5x-3y\right)^2-16\left(x^2-y^2\right)+1\)

\(=25x^2+9y^2-30xy-16x^2+16y^2+1\)

\(=9x^2-30xy+25y^2+1\)

\(=\left(3x-5y\right)^2+1>0\) \(\forall x;y\)

Giúp mk đi các bạn ơi

Mk nhớ ơn suốt đời

có \(x^2=y^2+4x^2\)

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

Tiếp tục với \(\left(5x-3y+8z\right)\left(5x-3y-8z\right)+1\)

\(=\left(5x-3y\right)^2-\left(8x\right)^2+1\)

\(=25x^2-30xy+9y^2-64x^2+1\)

\(=25x^2-30xy+9y^2-16\cdot4x^2+1\)

Thay \(x^2-y^2=4z^2\)

\(\Rightarrow25x^2-30xy+9y^2-16\cdot4x^2+1\)

\(=25x^2-30xy+9y^2-16\cdot\left(x^2-y^2\right)+1\)

\(=25x^2-30xy+9y^2-16x^2+16y^2+1\)

\(=9x^2-30xy+25y^2+1\)

\(=\left(9x^2-30xy+25y^2\right)+1\)

\(=\left(3x-5y\right)^2+1\)

ta có \(\left(3x-5y\right)^2\ge0\)

\(\Rightarrow\left(3x-5y\right)^2+1>0\)

\(\Rightarrow\left(5x-3x+8z\right)\left(5x-3y-8z\right)+1\)luôn dương với mọi x;y

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

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

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

Sửa đề: x² = y² + z²

=> z² = x² - y²

Ta có:

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

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

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

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

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

=> ĐPCM

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

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

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

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