\(f\left(x\right)=x^{10}+101x^9+101x^8-101x^7+...-101x+101\)

\(=x^{10}-100x^9-x^9+100x^8+x^8-100x^7-x^7+....-101x+101\)

\(=x^9.\left(x-100\right)-x^8\left(x-100\right)+x^7\left(x-100\right)-.....+x\left(x-100\right)-\left(x-101\right)\)

\(\Rightarrow f\left(100\right)=1\)

Ta có:

`101=100+1=x+1`

`⇒f(x)=x^10 - 101 x^9 +  ... -101x+101`

`⇒ f(100)=  x^10 - (x+1) x^9 + ... -(x+1).x+x+1`

              `=x^10 - x^10 - x^9 + ... -x^2 -x +x+1`

              `=1`

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 -x^{7 +.... -}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

Sai rồi bạn ơi, f(x) =x^10-101x^9+101x^8-101x^7+...-101x+''101'' chứ không phải 100 đâu bạn ơi!

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 hiểu tính f(100) thì tính f(x) tại x=100

Ta có:x+1=100+1=101

Thay vào biểu thức thì ta có:

\(f\left(x\right)=x^{10}-101x^9+101x^8-101x^7+...-101x+30\)

\(f\left(x\right)=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x+30\)

\(f\left(x\right)=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^2-x+30\)

\(f\left(x\right)=-x+30\)

\(f\left(100\right)=-100+30=-70\)

Vậy f(100)=-70

\(f\left(100\right)=100^8-101.100^7+101.100^6-101.100^5+...+101.100^2-101.100+25\)

=100⁸-(100+1).100⁷+(100+1).100⁶+...+(100+1).100²-(100+1).100+25

=100⁸-100⁸-100⁷+100⁷+100⁶+...+100³+100²-100²-100+25

=-75

Mơn Triều nhé!

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) -...+ x(x - 100) - (x - 25)

f(100) = 100⁷(100 - 100) - 100⁶(100 - 100) + 100⁵(100 - 100) -...+ 100(100 - 100) - (100 - 25)

f(100) = 0 - 0 + 0 -...+ 0 - 75

f(100) = -75

Ta có : với f(100) thì x = 100 ( điều đương nhiên)

=> 101 = x + 1

Thay vào f(x) , ta lại có :

\(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x+25\)

\(f\left(x\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x+25\)

\(f\left(x\right)=x^8-x^8-x^7+x^7+x^6-x^6-x^5+....+x^2+x+25\)

\(f\left(x\right)=x+25\)

\(f\left(100\right)=100+25=125\)

giúp chứng minh x=100 không là nghiệm của đa thức

\(f\left(x\right)=x^8-101x^7+101x^6-101x^5+...+101x^2-101x+25\)

\(f\left(x\right)=x^8-100x^7-x^7+100x^6+x^6-100x^5-x^5+...+100x^2+x-100x-x+25\)

\(f\left(x\right)=x^7\left(x-100\right)-x^6\left(x-100\right)+x^5\left(x-100\right)-...+x\left(x-100\right)-x+25\)

\(f\left(100\right)=x^7.0-x^6.0+x^5.0-...+x.0-100+25\)

\(f\left(100\right)=25-100=-75\)

\(f\left(100\right)=100^8-101.\left(100^7-100^6+100^5-100^4+100^3-100^2+100\right)+25\)

ta có: \(100.\left(100^7-100^6+...-100^2+100\right)+\left(100^7-100^6+...-100^2+100\right)=100^8+100\)

\(\Rightarrow f\left(x\right)=100^8-100^8-100+25=-75\)

p/s: cách này ngắn nên tớ làm, ko phải câu (có 1 bạn tl rồi)