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

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

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

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

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`

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

=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)

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

Mk ko hiểu lắm! Hoàng Phúc

-x+25=25?

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:\(101=100+1=x+1\)

\(\Rightarrow F\left(100\right)=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+x+1\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1=1\)