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Ta có: x=100
nên x+1=101
Ta có: \(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x^2-101x+25\)
\(=x^8-x^7\left(x+1\right)+x^6\left(x+1\right)-x^5\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+25\)
\(=x^8-x^7-x^7+x^7+x^6-x^6-x^5+x^5-x^4+...+x^3+x^2-x^2-x+25\)
\(=-x+25\)
\(=-100+25=-75\)
Ta có: x=100
\(\Leftrightarrow x+1=101\)
Ta có: \(f\left(x\right)=x^{10}-101x^9+101x^8-101x^7+...+101x+2021\)
\(=x^{10}-x^9\cdot\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x\left(x+1\right)+2021\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^2+x+2021\)
\(=x+2021\)
\(=100+2021=2121\)
\(f\left(100\right)\Rightarrow x=100\)
\(\Rightarrow x+1=101\)
Thay x + 1 = 101 ta được:
\(f\left(100\right)-x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^2-\left(x+1\right)x+25\)
\(=x^8-\left(x^8+x^7\right)+\left(x^7+x^6\right)-\left(x^6+x^5\right)+...+\left(x^3+x^2\right)-\left(x^2+x\right)+25\)
\(=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...+x^3+x^2-x^2-x+25\)
\(=-x+25\)
\(=-100+25\)
\(=-75\)
\(f\left(x\right)=x^8-101x^7+101x^6-...-101x+25\)
\(f\left(100\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-...-\left(x+1\right)x+25\)
\(f\left(100\right)=x^8-x^8-x^7+x^7+x^6-...-x^2-x+25\)
\(f\left(100\right)=-x+25=-100+25=-75\)
Bài 1:
\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)
\(f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\)
Do đó: \(2\cdot f\left(x\right)=10x^4-6x^3+4x^2+8x-14\)
=>\(f\left(x\right)=5x^4-3x^3+2x^2+4x-7\)
\(g\left(x\right)=5x^4-3x^3+2x^2+4x-7-4x^4+6x^3-7x^2-8x+9\)
\(=x^4+3x^3-5x^2-4x+2\)
f(100)=x8-(100+1)x7+(100+1)x6-(100+1)x5+....+(100+1)x2-(100+1)x+25
=x8-(x+1)x7+(x+1)x6-(x+1)x5+....+(x+1)x2-(x+1)x+25
=x8-x8-x7+x7+x6-x6-x5+...+x3+x2-x2-x+25
=25
vậy f(100)=25
f(100)=1008-101.1007+101.1006-101.1005+...+101.1002-101.100+5
=1008-(100+1)1007+(100+1)1006-(100+1)1005+...+(100+1)1002-(100+1)100+25
=1008-1008-1007+1007+1006-1006-1005+...+1003+1002-1002-100+25
=0-100+25=-75