Question

[vũ thị vân anh]

x⁸ + x⁶  + x⁴ + x²  +1 

giúp mk vs . help me !!!!!!!!!

Answer 1

[Phạm Thành Đông]

Đặt \(A=x^8+x^6+x^4+x^2+1\)

\(A=x^4\left(x^4+x^2+1+\frac{1}{x^2}+\frac{1}{x^4}\right)\)

\(A=x^4\left[\left(x^4+\frac{1}{x^4}\right)+\left(x^2+\frac{1}{x^2}\right)\right]\)

Đặt \(x^2+\frac{1}{x^2}=a\)thì \(\left(x^2+\frac{1}{x^2}\right)=a^2\Rightarrow x^4+2+\frac{1}{x^4}=a^2\Rightarrow x^4+\frac{1}{x^4}=a^2-2\), lúc đó:

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

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

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

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

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

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

\(A=x^4\left(x^2+\frac{1}{x^2}-1\right)\left(x^2+\frac{1}{x^2}+2\right)\)

\(A=\left[x^2\left(x^2+\frac{1}{x^2}-1\right)\right]\left[x^2\left(x^2+\frac{1}{x^2}+2\right)\right]\)

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

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

Answer 2

[nguyễn ngọc linh]

=1 nha bn