Tính giá trị của biểu thức:
B=x2006-8x2005+8x2004-...+8x2-8x-5 tại x=7
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x=4
=>x+1=5
A=(x+1)x^5 -(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-1
=x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1
=x^6-x-1
=4^6-4-1
=4091
\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)
\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)
a)
\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)
\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)
\(P\left(9\right)=1\)
b)
\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)
\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)
\(Q\left(7\right)=2\)
x=7 nên x+1=8
\(B=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-x^{13}+x^{13}-...-x^3-x^2+x^2+x+5\)
=x+5
=7+5
=12
\(A=x^5-5x^4+5x^3-5x^2+5x-6\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x-2\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-2\)
\(=-2\)
ta có: 8=7+1=x+1
\(B=x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\)
\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(\Rightarrow B=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)
\(\Rightarrow B=x-5\)
\(\Rightarrow B=7-5\)
\(\Rightarrow B=2\)
B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5
B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5
B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5
B = 7-5=2
Tham khảo cách này nhoá~
\(x=7\Rightarrow8=x+1\left(1\right)\)
Thay \(1\) vào \(F\) ta có:
\(F=x^{2006}-\left(x+1\right)^{2005}+\left(x+1\right)^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\)
\(F=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\)
\(F=-7-5\)
\(\Rightarrow F=-12\)