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a: \(=4xy\left(1-5x^2y\right)\)
b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)
d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)
a, \(x^4+2x^2+1-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+x+1\right)\left(x^2-x+1\right)\)
b, \(x^4+x^2+1\)
= \(x^4+2x^2+1-x^2\)
= .. ( như phần a )
c, \(y^4+64\)
= \(\left(y^2+8\right)\left(y^2-8\right)\)
d, \(4xy+3z-12y-xz\)
\(=4y\left(x-3\right)-z\left(x-3\right)\)
\(=\left(x-3\right)\left(4y-z\right)\)
e, \(x^2-4xy+4y^2-z^2+6z-9\)
\(=\left(x-2y\right)^2-\left(z-3\right)^2\)
g, \(x^2-4xy+5x+4y^2-10y\)
\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)
\(=\left(x-2y\right)^2+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+5\right)\)
h, \(x^2-7x+6\)
\(=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
i, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
Bạn xem lại đề nhé: Ví dụ chọn x=2, y=1 ta có: 22-4.2.1+1+2=-1<0
\(x^2+4xy-4z^2+4y^2\)
\(=x^2+4xy+4y^2-4z^2\)
\(=\left(x+2y\right)^2-4z^2\)
\(=\left(x+2y-2z\right)\left(x+2y+2z\right)\)
\(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-16\)
\(=\left(x+1-4\right)\left(x+1+4\right)\)
\(=\left(x-3\right)\left(x+5\right)\)
x2 - 2xy + 2y2 - 2x + 6y + 13 = 0
<=> x2 - 2x(y + 1) + 2y2 + 6y + 13 = 0
<=> x2 - 2x(y + 1) + (y + 1)2 + y2 + 4y + 12 = 0
<=> (x - y - 1)2 + (y + 1)2 + (y + 2)2 + 8 = 0
Vô lí do VT > 0 vs mọi x; y
=> Ko tìm đc gtri của N