x²(x+1)(x³-X²-1)

\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)

a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)

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

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

a: Ta có: \(x^3-2x^2y-4x+8y\)

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

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

b: Ta có: \(a^2x^2-a^2y^2-b^2x^2+b^2y^2\)

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

\(=\left(x-y\right)\left(x+y\right)\left(a-b\right)\left(a+b\right)\)

a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

a) \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

b) Ta có: \(4x^4+y^4\)

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

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

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

a, \(x^6-x^4+2x^3+2x^2\)

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

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

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

b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)

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

x⁴-2x³+2x-1

=(x⁴-1)+(-2x³+2x)

=(x²+1)(x²-1)-2x(x²-1)

=(x²-1)(x²+1-2x)

=(x-1)(x+1)(x-1)²

=(x-1)³(x+1)