Question

A=x^6-x^5(x-1)-x^4(x+1)+x^3(x-1)+x^2(x+1)-x(x-1)+1 với x=999

Answer 1

\(A=x^6+x^5\left(x-1\right)-x^4\left(x+1\right)+x^3\left(x-1\right)+x^{20}\left(x+1\right)-x\left(x-1\right)+1\)

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

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

\(A=x+1\) x = 999

=> A = 999 + 1 = 1000

Answer 2

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

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

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

\(A=x+1\)

Thay \(x=999\)vào A, ta có:

\(A=x+1=999+1=1000\)

Vậy tại \(x=999\)thì \(A=1000\)

Answer 3

\(A=1000\)

hok tốt ~!!!