Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1, <=> \(\left(4x\right)^2-\left(9y\right)^2\)=\(\left(4x-9y\right)\left(4x+9y\right)\)
1) \(16x^2-81.y^2=\left(4x\right)^2-\left(9.y\right)^2=\left(4x-9y\right)\left(4x+9y\right)\)
2) \(\left(5x-3y\right)^2-\left(3x-5y\right)^2=\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)=\left(2x+2y\right).\left(8x-8y\right)\)
\(=16.\left(x+y\right)\left(x-y\right)\)
3)\(4x^2-y^2+4y-4=4x^2-\left(y^2-4y+4\right)=\left(2x\right)^2-\left(y-2\right)^2=\left(2x-y+2\right).\left(2x+y-2\right)\)
4)\(9.\left(x-y\right)^2-16.\left(2x+y\right)^2=3^2.\left(x-y\right)^2-4^2.\left(2x+y\right)^2=\left(3x-3y\right)^2-\left(8x+4y\right)^2\)
\(=\left(3x-3y-8x-4y\right)\left(3x-3y+8x+4y\right)=\left(-5x-7y\right).\left(11x+y\right)\)
a/ 4x^2 + 4x +1=(2x)2+2.2x.1+12=(2x+1)2=(2x+1)(2x+1)
c/ 81y^4 - 16x^6=(9y2)2-(4x3)2=(9y2+4x3)(9y2-4x3)
d/ 4x^2 + y^2 + z^2 + 4xy + 2yz + 4xz=[(2x)2+4xy+y2]+(4xz+2yz)+z2
=(2x+y)2+2z(2x+y)+z2
=(2x+y+z)2
=(2x+y+z)(2x+y+z)
\(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x-y\right)^2-4\left(4x-y\right)\)
\(=\left(4x-y\right)\left(4x-y-4\right)\)
a) \(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x\right)^2-8xy+y^2+4\left(y-4x\right)\)
\(=\left(4x-y\right)^2+4\left(y-4x\right)\)
\(=\left(y-4x\right)^2+4\left(y-4x\right)=\left(y-4x\right)\left(y-4x+4\right)\)
