Bài 1:
`f(x) = (6x^4 -3x^2 -5 + 4x^4 - 6x^3 + 7x^2 +8x-9):2`
`f(x)= [(6x^4 + 4x^4) - (3x^2 - 7x^2) - ( 5+9) - 6x^3)]:2`
`f(x) = (10x^4 + 4x^2 - 14 - 6x^3):2`
`f(x)= 5x^4 + 2x^2 - 7 - 3x^3`
Vì `f(x) + g(x)= 6x^4 - 3x^2-5`
`=> g(x) = 6x^4 - 3x^2 -5 -f(x)`
`= 6x^4 - 3x^2 -5 - (5x^4 + 2x^2 - 7 - 3x^3)`
`= 6x^4 - 3x^2 -5 - 5x^4 - 2x^2 + 7 + 3x^3`
`= (6x^4 - 5x^4) - (3x^2 + 2x^2) - (5 -7) + 3x^3`
`= x^4 - 5x^2 + 2 + 3x^3`
Vậy `f(x)= 5x^4 + 2x^2 - 7 - 3x^3`
`g(x) =x^4 - 5x^2 + 2 + 3x^3`
Bài 2:
Ta có:
`f(x) - g(x)= (x^(2n) - x^(2n-1) +...+x^2 -x +1) - ( -x^(2n+1) +x^(2n) - x^(2n-1) +...+x^2 -x+1)`
`= x^(2n) - x^(2n-1) +...+x^2 -x +1 + x^(2n+1) - x^(2n) + x^(2n-1) -....-x^2 + x-1`
`= (x^(2n) - x^(2n) ) - (x^(2n-1) - x^(2n-1) ) +.....+ (x^2 - x^2) - (x -x) +(1-1) + x^(2n+1)`
`= x^(2n+1)`
Thay `x = 1/10` vào biểu thức trên ta được:
`(1/10)^(2n+1) = 1^(2n+1)/(10^(2n+1)) = 1/(10^(2n+1))`
Vậy `f(x) - g(x)=1/(10^(2n+1))`
Bài 3:
Tính `f(100) => x= 100 = x -100=0`
Ta có: `f(x)= x^8 - 101 x^7 + 101x^6 +...+ 101x^2 -101x+25`
`f(x) = x^8 - (100+1)x^7 +(100+1)x^6 +...+ (100+1)x^2 - (100+1)x + 25`
`f(x) = x^8 - 100x^7 - x^7 + 101x^6 + x^6 +...+ 100x^2 + x^2 -100x -x +25`
`f(x) = x^7(x-100) - x^6( x -100) + ....+ x(x -100) - x +25`
`f(x) = x^7 . 0 - x^6 .0 +...+ x.0 - x+25`
`=> f(100) = -100+25`
`=> f(100) = -75`
Vậy `f(100) = -75`
