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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
1. Ta có :
f(x) = ( m - 1 ) . 12 - 3m . 1 + 2 = 0
f(x) = m - 1 - 3m + 2 = -2m + 1 = 0
\(\Rightarrow m=\frac{1}{2}\)
2.
a) M(x) = -2x2 + 5x = 0
\(\Rightarrow-2x^2+5x=x.\left(-2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\-2x+5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}}\)
b) N(x) = x . ( x - 1/2 ) + 2 . ( x - 1/2 ) = 0
N(x) = ( x + 2 ) . ( x - 1/2 ) = 0
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-\frac{1}{2}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2}\end{cases}}\)
c) P(x) = x2 + 2x + 2015 = x2 + x + x + 1 + 2014 = x . ( x + 1 ) + ( x + 1 ) + 2014 = ( x + 1 ) . ( x + 1 ) + 2014 = ( x + 1 )2 + 2014
vì ( x + 1 )2 + 2014 > 0 nên P(x) không có nghiệm
Ta có : x=100=>101=x+1
Thay vào f(x), ta được : x10 -(x+1)x9 +(x+1)x8 - (x+1)x7 +....-(x+1)x +100
<=> x10 - x10 -x9 +x9 + x8 -x8 -x7 +.... -x2 -x +100
<=> -x+100
=> f(100) = -x+100=-100+100=0
ta có :
\(f\left(x\right)=x^{10}-101x^9+101x^8-...-101x+101\)
\(=x^{10}-x^9-100x^9+x^8+100x^8-...-x-100x+100+1\)
ta có :
\(f\left(100\right)=100^{10}-100^9-100\times100^9+100^8+100\times100^8-...-100-100\times100+100+1\)
\(=100^{10}-100^{10}-100^9+100^9+100^8-...-100^2-100+100+1\)
\(=1\)
vậy f(100)=1
\(f\left(100\right)\Rightarrow x=100\)
\(\Rightarrow x+1=101\)
Thay x + 1 = 101 ta được:
\(f\left(100\right)-x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^2-\left(x+1\right)x+25\)
\(=x^8-\left(x^8+x^7\right)+\left(x^7+x^6\right)-\left(x^6+x^5\right)+...+\left(x^3+x^2\right)-\left(x^2+x\right)+25\)
\(=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...+x^3+x^2-x^2-x+25\)
\(=-x+25\)
\(=-100+25\)
\(=-75\)
