Question

a,. x^3+1–x^2–x

b,,. x^4-1-3(x^2+1)

c,,,x^2+y^2-2xy-4z^2

d,, x^2-4x+4-y^2-6y-9

e,,, (x^2-2x+1)^3-y^6

f,, (x+y)^3-x^3-y^3

Phân tích các đa thức sau thành nhân tử bằng hằng đẳng thức

Answer 1

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

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

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

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

c/ \(x^2+y^2-2xy-4z^2=\left(x-y\right)^2-\left(2z\right)^2\)

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

Answer 2

d/ \(x^2-4x+4-\left(y^2+6y+9\right)\)

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

\(=\left(x+y+1\right)\left(x-y-5\right)\)

e/\(\left(x^2-2x+1\right)^3-\left(y^2\right)^3\)

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

\(=\left[\left(x-1\right)^2-y^2\right]\left[\left(x-1\right)^4+y^4+2\left(x-1\right)^2y^2-\left(xy-y\right)^2\right]\)

\(=\left(x+y-1\right)\left(x-y-1\right)\left[\left(\left(x-1\right)^2+y^2\right)^2-\left(xy-y\right)^2\right]\)

\(=\left(x+y-1\right)\left(x-y-1\right)\left[\left(x-1\right)^2+y^2-xy+y\right]\left[\left(x-1\right)^2+y^2+xy-y\right]\)

f/ \(\left(x+y\right)^3-\left(x^3+y^3\right)\)

\(=x^3+y^3+3xy\left(x+y\right)-\left(x^3+y^3\right)\)

\(=3xy\left(x+y\right)\)