Ta có x =7
=>x+1=8
\(\Rightarrow\)\(A=x^{15}-8x^{14}+8x^{13}-8x^{12}+.......8x^2+8x-5\)
\(\Rightarrow x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...\left(x+1\right)x^2\)
\(+\left(x+1\right)x^5\)
\(\Rightarrow x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x-5\)
\(\Rightarrow x-5\Leftrightarrow A=7-5=2\Rightarrow A=2\)
Vậy A=2 khi x=7
Vì x=7 nên 8 = x + 1
Thay 8 = x + 1 vào biểu thức A ta có
\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+....-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+....-x^3-x^2+x^2+x-5\)
\(=x-5\)
Mà x = 7
Nên \(A=7-5\)
\(=2\)
Vậy A = 2 tại x=7
