Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+x-2=\left(x-2\right)\left(x^2+x+1\right)\)
\(3\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2-x^2-x\right)\)
x^4 + x^2 + 1
= x^4 + 2x^2 + 1 - x^2
= ( x^2 + 1)^2 - x^2
= ( x^2 - x + 1 )( x^2 + x + 1)
\(x^2+x-1\)=\(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{5}{4}\)=\(\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2\)=\(\left(x+\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x+\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)
x4+x2+1
=x4-x+x2+x+1
=x(x3-1)+(x2+x+1)
=x(x-1)(x2+x+1)+(x2+x+1)
=(x2-x)(x2+x+1)+(x2+x+1)
=(x2+x+1)(x2-x+1)
\(x^7+x^5+x^4+x^3+x^2+1\)
\(=x^7+x^6-x^6-x^5+2x^5+2x^4-x^4-x^3+2x^3+2x^2-x^2-x+x+1\)
\(=\left(x^7+x^6\right)-\left(x^6+x^5\right)+\left(2x^5+2x^4\right)-\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(x^2+x\right)+\left(x+1\right)\)
\(=x^6.\left(x+1\right)-x^5.\left(x+1\right)+2x^4\left(x+1\right)-x^3\left(x+1\right)+2x^2\left(x+1\right)-x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^6-x^5+2x^4-x^3+2x^2-x+1\right)\)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
KHÔNG BÍT