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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(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
Ta có: 101 = 100+1=x+1
Khi đó :
\(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x^2-101x+25\)
\(f\left(x\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+.....+\left(x+1\right)x^2-\left(x+1\right)x+25\)
\(f\left(x\right)=x^8-x^8-x^7+x^7+x^6-x^6+x^5+...+x^3+x^2-x^2-x+25\)
\(f\left(x\right)=-x+25\)
Vậy \(f\left(100\right)=-100+25=-75\)
f(100)=> x=100
=>x+1=101
thay x+1=101 ta được:
f(100)=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
=x8-x8-x7+x7+x6-x6-x5+...+x3+x2-x2-x+25
=-x+25
=-100+25
=-75
f(x) = x8 - 101x7 + 101x6 - 101x5 + ... + 101x2 - 101x + 25
f(x) = x8 - ( 100x7 + x7 ) + ( 100x6 + x6 ) - ( 100x5 + x5 ) + ... + ( 100x2 + x2 ) - ( 100x + x ) + 25
f(x) = x7 . ( x - 100 ) - x6 . ( x - 100 ) - x5 . ( x - 100 ) - x4 . ( x - 100 ) + ... + x . ( 100 - x ) - ( x - 25 )
nên f(100) = - ( 100 - 25 ) = -75