Question

\(B=x^{15}-8x^{14}+8x^{13}-8x^2+...-8x^2+8x\)\(-5\)

tại x=7

Answer 1

khó thế

Answer 2

Bằng 2 bạn ạ Thay 8=7+1= x+1 Sau đó tính lần lượt ra

Answer 3

Vì x = 7

\(\Rightarrow\)\(x+1=8\)

\(\Rightarrow\)\(A=x^{15}\)\(-\)\(8x^{14}\)\(+\)\(8x^{13}\)\(-\)\(8x^{12}\)\(+\)... \(-\)\(8x^2\)\(+8x-5\)

\(=\)\(x^{15}\)\(-\left(x+1\right)x^{14}\)\(+\left(x+1\right)x^{13}\)\(-\left(x+1\right)x^{12}\)\(+\)... \(-\)\(\left(x+1\right)x^2\)\(+\left(x+1\right)x-5\)

\(=\)\(x^{15}\)\(-\)\(x^{15}\)\(-\)\(x^{14}\)\(+\)\(x^{14}\)\(+\)\(x^{13}\)\(-\)\(x^{13}\)\(-\)\(x^{12}\)\(+\)... \(-\)\(x^3\)\(-\)\(x^2\)\(+\)\(x^2\)\(+\)\(x\)\(-\)\(5\)

\(\Rightarrow\)\(x=-5\)

\(\Rightarrow\)\(A=7-5=2\)

Vậy  \(A=2\) khi  \(x=7\)