\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2zt\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2zt-x^2-y^2+z^2+t^2\right)\left(2xy+2zt+x^2+y^2-z^2-t^2\right)^2\)

Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2tz-x^2-y^2+z^2+t^2\right)\left(2xy+2tz+x^2+y^2-z^2-t^2\right)\)

\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)

\(=\left(z+t-x+y\right)\left(z+t+x-y\right)\left(x+y-t+z\right)\left(x+y+t-z\right)\)

\(1)x^3-x^2y-4x-4y=x^2\left(x-y\right)-4\left(x-y\right)=\left(x^2-2^2\right)\left(x-y\right)=\left(x^2-4x+4\right)\left(x-y\right)\)

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

\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)

A = -(x-y)-z(x-y)=(x-y)(-1-z)=(y-x)(z+1)

A = -x - z(x - y) + y

A = -x - zx + zy + y

A = -(-x - zx + zy + y)

A = x + zx - zy - y

A = x + zx - y - zy

A = x(1 + z) - y(1 + z)

A = (x - y)(1 + z)

\(\left(x^2-xy+y^2\right)^2\left(x^2+xy+y^2\right)^2\)

Phương trình thuần nhất đẳng cấp bậc 8 bạn nha :D

\(4(x^2y^2+z^2t^2+2xyzt)-(x^2+y^2-z^2-t^2)^2\)

\(=[2(xy+zt]^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt)^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt-x^2-y^2+z^2+t^2)(2xy+2zt+x^2+y^2-z^2-t^2)^2\)

\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)

\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)

\(\left(x+y\right)^2+3\left(x+y\right)-10\)

\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)

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

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

Ta có:

 (x + y + z)² + (x + y – z)² – 4z²

^{\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)}

​\(=\left(x+y-z\right)\left(x+y+3z+x+y-z\right)\)​

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

\(=x^2+y^2+z^2+2xy+2yz+2xz+x^2+y^2+z^2+2xy-2xz-2yz-4z^2\)

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

\(=2\left(x^2+2xy+y^2-z^2\right)\)

\(=2\left(x+y-z\right)\left(x+y+z\right)\)

\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)

\(x^3+x^2y-4x-4y\)

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

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

\(=\left(x+y\right)\left(x+2\right)\left(x-2\right)\)

Làm:

x³+x²y-4x-4y

= x²(x+y)-4(x+y)

= (x+y)(x²-4)

= (x+y)(x-2)(x+2)

Kl:.........