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

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

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

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

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

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

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

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

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

Giúp mk đi các bạn ơi

Mk nhớ ơn suốt đời

Áp dụng hằng đẳng thức \(\left(a-b\right).\left(a+b\right)=a^2-b^2\) vào ta được:

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

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

Ta dùng tính chất:

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

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

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

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

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

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

Ta có: \(\left(3x-5y\right)^2+1\ge0\) \(\forall x,y.\)

\(\Rightarrow\left(3x-5y\right)^2\) luôn dương.

\(\Rightarrow\left(5x-3y+8z\right).\left(5x-3y-8z\right)+1\) luông dương \(\forall x,y\left(đpcm\right).\)

Chúc bạn học tốt!

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

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:

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

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

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

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

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