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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
Ta có:
\(F\left(100\right)=100^{10}-101.100^9+101.100^8-101.100^7+...-101.100+101\)
\(=100-\left(100+1\right).100^9+\left(100+1\right).100^8-\left(100+1\right).100^7+...-\left(100+1\right).100+101\)
\(=100^{10}-100^{10}-100^9+100^9+100^8-100^8-100^7+...-100^2-100+101\)
\(=1\)
Ta có:\(101=100+1=x+1\)
\(\Rightarrow F\left(100\right)=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+x+1\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1=1\)
x=100
nên x+1=101
\(f\left(x\right)=x^{2014}-\left(x+1\right)\left(x^{2013}-x^{2012}+...-x^2+x\right)+25\)
\(=x+25\)
=x+25=100+25=125
\(f\left(x\right)=x^8-101x^7+101x^6-...-101x+25\)
\(f\left(100\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-...-\left(x+1\right)x+25\)
\(f\left(100\right)=x^8-x^8-x^7+x^7+x^6-...-x^2-x+25\)
\(f\left(100\right)=-x+25=-100+25=-75\)
Vì \(VT\ge0\Leftrightarrow VP\ge0\Leftrightarrow101x\ge0\Leftrightarrow x\ge0\)
=>\(\left|x+\frac{1}{101}\right|=x+\frac{1}{101};\left|x+\frac{2}{101}\right|=x+\frac{2}{101};...;\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
=>\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)
=>\(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=100x\)
=>\(100x+50=101x\)
=> x = 50
bài này khó quá ta
mk chịu mất rùi
chúc bn học gioi!
và các bn khác giúp bn này nha'
hihi@@@
Ta có:
\(x^{10}-\left(100+1\right)x^9+\left(100+1\right)x^8-\left(100+1\right)x^7+.....-\left(100+1\right)x+100+1\)
\(=x^{10}-100x^9-x^9+100x^8+x^8-100x^7-x^7+......-100x-x+100+1\)
\(f\left(x\right)=x^{10}-\left(100+1\right)x^9+\left(100+1\right)x^8-\left(100+1\right)x^7+...-\left(100+1\right)x+100+1\)
\(=x^{10}-100x^9-x^9+100x^8+x^8-100x^7-x^7+...-100x-x+100+1\)
\(=x^9\left(x-100\right)-x^8\left(x-100\right)+x^7\left(x-100\right)-...+x\left(x-100\right)-\left(x-100\right)+1\)
\(=\left(x-100\right)\left(x^9-x^8+x^7-...+x-1\right)+1\)
Ta có: \(f\left(100\right)=\left(100-100\right)\left(100^9-100^8+100^7-...+100-1\right)+1\)
\(=0+1=1\)
Vậy f(100) = 1.
Ta có:\(f\left(x\right)=x^8-100x^7-x^7+100x^6-....+x^2-100x-x+100-75\)
\(=x^7\left(x-100\right)-x^6\left(x-100\right)-....+x\left(x-100\right)-\left(x-100\right)-75\)
Nên \(f\left(100\right)=x^7.\left(100-100\right)-x^6\left(100-100\right)-....+x\left(100-100\right)-\left(100-100\right)-75\)
\(=-75\)
Với x= 100 thì 101=x+1 nên ta có f(100)=x\(^8\)-(x+1)x\(^7\)=(x+1)x\(^6\)-(x+1)x\(^5\)+....-(x+1)+25=x\(^8\)-x\(^8\)+x\(^7\)-......-x-1+25=24