2. cho f(x)= x8-101x7+101x6-101x5+...+ 101x2-101x+25. tính f(100)?
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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
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
\(x=100\Rightarrow x+1=101\)
\(f\left(x\right)=x^8-\left(x+1\right).x^7+\left(x+1\right).x^6-\left(x+1\right).x^5+....+\left(x+1\right).x^2+\left(x+1\right).x+25\)
\(f\left(x\right)=x^8-x^8-x^7+x^7+x^6-x^6-x^5+.....+x^3+x^2-x^2+x+25\)
\(f\left(100\right)=100+25=125\)
Vì x=100 nên x+1=101
Thay 101=x+1 vào f(x) ta có:
f(x) = 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
= x+25
=> f(100) = 100+25=125
Vậy f(100) = 125
Cách2:
f(100) = 1008 - 101.1007 + 101.1006-101.1005+...+101.1002-101.100+25
=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+100 -1002-100+25
=-100+25=75
Vậy f(100)=75