Question

tính GTBT :

a. A=x²⁰-2006.x¹⁹+2006.x¹⁸-2006.x¹⁷+.....+2006.x²-2006.x+2006

tại x=2005

b. B=x⁵-15.x⁴+16.x³-29.x²+13.x

tại x=14

Answer 1

a,

x=2005=> 2006=x+1 . Thay vào biểu thức A có:

\(A=x^{20}-\left(x+1\right)x^{19}+\left(x+1\right)x^{18}-\left(x+1\right)x^{17}+....+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)A=\(x^{20}-x^{20}+x^{19}-x^{19}+x^{18}-x^{18}+...+x^3+x^2-x^2-x+x+1\)

A=1

Answer 2

b,

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

B=\(x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)

B=x=14