Question

Tính giá trị của biểu thức: 

Answer 1

Ta có: 19=18+1=x+1

\(D=x^{12}-19x^{11}+19x^{10}+...+19x^2-19x+1\)

\(\Rightarrow D=x^{12}-\left(x+1\right)x^{11}+\left(x+1\right)x^{10}+...+\left(x+1\right)x^2-\left(x+1\right)x+1\)

\(\Rightarrow D=x^{12}-x^{12}-x^{11}+x^{11}+x^{10}+...+x^3+x^2-x^2-x+1\)

\(\Rightarrow D=-x+1\\ \Rightarrow D=-18+1\\ \Rightarrow D=-17\)

Answer 2

x=18

nên x+1=19

\(D=x^{12}-x^{11}\left(x+1\right)+x^{10}\left(x+1\right)-x^9\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+1\)

\(=x^{12}-x^{12}-x^{11}+x^{11}+x^{10}-x^{10}+...+x^3+x^2-x^2-x+1\)

=-x+1

=-18+1=-17