chứng minh rằng : (5x-3y+4z)(5x- 3y-4z)=(3x-5y)2 nếu x2=y2+z2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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 \(x^2-y^2-z^2=0\)
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
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:
(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