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

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

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

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

\(x^4+2x^3+x^2+x+1\)

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

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

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

\(x^4-2x^3+2x-1\)

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

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

\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

\(= \)\(5x^2-4x^2+8x-4-5\)

\(=\)\(x^2+8x-9\)

\(=x^2+9x-x-9\)

\(=(x-1)(x+9)\)

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

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

a)  x^6 - x^4 + 2x^3 + 2x^2 

=x²(x⁴-x²+2x+2)

=x²[x⁴-2x³+2x²+2x³-4x²+4x+x²-2x+2]

=x²[x²(x²-2x+2)+2x(x²-2x+2)+(x²-2x+2)

=x²[(x²+2x+1²)(x²-2x+2)]

=x²(x+1)²(x²-2x+2)

b) x^(m+4) + x^(m+1) - x - 1

Ta thấy x=-1 là nghiệm của đa thức

=>đa thức có 1 hạng tử là x+1

=>đa thức đc phân tích là

=(x+1)(x^m+3-x^m+2+x^m+1-1)

\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2=a^2+b^2+a^2b^2+1=a^2\left(b^2+1\right)+\left(b^2+1\right)=\left(a^2+1\right)\left(b^2+1\right)\)

\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2=a^2b^2+a^2+b^2+1=\left(a^2b^2+a^2\right)+\left(b^2+1\right)=a^2\left(b^2+1\right)+\left(b^2+1\right)=\left(a^2+1\right)\left(b^2+1\right)\)

\(\left(ab-1\right)^2+\left(a+b\right)^2\)

\(=a^2b^2+1+a^2+b^2\)

\(=\left(a^2+1\right)\left(b^2+1\right)\)

a) \(\frac{1}{4}x^2-5xy+25y^2=\left(\frac{1}{2}x\right)^2-5xy+\left(5y\right)^2\)

\(=\left(\frac{1}{2}x-5y\right)^2\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4+2x+1\right)\times\left(7x-4-2x-1\right)=\left(9x-3\right)\times\left(5x-5\right)\)

\(=3\times5\times\left(3x-1\right)\times\left(x-1\right)=15\times\left(3x-1\right)\times\left(x-1\right)\)

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

d) \(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right)\left(25+5x^2+x^4\right)\)

\(x^4+x^3+2x^2+x+1\)

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

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

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

x^4+x^3+2x^2+x+1

=(x^4+2x^2+1)+(x^3+x)

=(x^2+1)^2+x(x^2+1)

=(x^2+1)(x^2+x+1)