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

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

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

Ủng hộ nha ^ _ ^

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

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

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

x⁸ + x +1=  x⁸  +x⁷ - x⁷ + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ -x⁴ +x³ -x³ + x² -x²  +x +1 

             =   (x²+x+1)*(x⁶ -x⁵+x³-x²+1)

x-x8+1+=121Vay X=112

Đa thức có dạng  \(x^{3a+1}+x^{3b+2}+1\)  thì đưa về dạng  \(\left(x^2+x+1\right)\cdot P\left(x\right)\) bạn nhé!

Bài làm:

\(x^5+x+1\)

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

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

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

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

\(x^5+x+1=x^5-x^2+x^2+x+1\)

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

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

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

\(x^2-5x+6\)

\(=x^2-5x+\frac{25}{4}-\frac{1}{4}\)

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

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

\(=\left(x-3\right)\left(x-2\right)\)

\(x^2-5x+6 \)

= \(x^2-2x-3x+6\)

= \(\left(x^2-2x\right)-\left(3x-6\right)\)

= \(x\left(x-2\right)-3\left(x-2\right)\)

= \(\left(x-2\right)\left(x-3\right)\)

\(6x^3+x^2-2x\)

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

a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)

b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)

c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)

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

tik nhé

​

\(x^{12}-3x^6+1=\left(x^{12}+x^9-x^6\right)-\left(x^9-x^3+x^6\right)-\left(x^3-1+x^6\right)=x^6\left(x^6+x^3-1\right)-x^3\left(x^6+x^3-1\right)-\left(x^6+x^3-1\right)\)

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

x¹²-2x⁶+1-x⁶

=(x⁶-1)²-x⁶

= (x⁶-1-x³)(x⁶-1+x³)

Ta có \(x^2-\left(m+n\right)x+m.n=\left(x^2-mx\right)-\left(nx-m.n\right)\)

\(=x\left(x-m\right)-n\left(x-m\right)=\left(x-m\right)\left(x-n\right)\)