1. 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: \(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x^2-101x+25\)
\(=x^8-x^7\left(x+1\right)+x^6\left(x+1\right)-x^5\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+25\)
\(=x^8-x^7-x^7+x^7+x^6-x^6-x^5+x^5-x^4+...+x^3+x^2-x^2-x+25\)
\(=-x+25\)
\(=-100+25=-75\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x=100
\(\Leftrightarrow x+1=101\)
Ta có: \(f\left(x\right)=x^{10}-101x^9+101x^8-101x^7+...+101x+2021\)
\(=x^{10}-x^9\cdot\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x\left(x+1\right)+2021\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^2+x+2021\)
\(=x+2021\)
\(=100+2021=2121\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=100\Rightarrow x+1=101\)
\(f\left(x\right)=x^8-\left(x+1\right).x^7+\left(x+1\right).x^6-\left(x+1\right).x^5+....+\left(x+1\right).x^2+\left(x+1\right).x+25\)
\(f\left(x\right)=x^8-x^8-x^7+x^7+x^6-x^6-x^5+.....+x^3+x^2-x^2+x+25\)
\(f\left(100\right)=100+25=125\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. cho f(x)=x8-101x7+101x6-101x5+...+101x2-101x+25. tính f(100)
f(x)=x8-101x7+101x6-101x5+...+101x2-101x+25
=x8-(100+1)x7+(100+1)x6-(100+1)x5+...+(100+1)x2-(100+1)x+25
f(100 ) hay x= 100
Thay 100 = x ,có :
=x8-(x+1)x7+(x+1)x6-(x+1)x5+...+(x+1)x2-(x+1)x+25
= x8 - x8 - x7+ x7 + x6 - x6 - x5 + x5 + .......................+ x3 + x2 - x2 + x + 25
= x+ 25
f(100 0 = 100 + 25 = 125
Vậy f(100 ) =125