`@` `\color{Silver}{\text{πk2}`

---------------------
`\color{red}{\text{c)}` `C=x^4+10x^3+10x^2+10x+10`

Ta có: `x=-9⇒x-1=-10⇒1-x=10`

`⇒C=x^4+(1-x)·x^3+(1-x)·x^2+(1-x)·x+(1-x)`

`⇒C=x^4+x^3-x^4+x^2-x^3+x-x^2+1-x`

`⇒C=1`

Vậy `C=1`
~~~~~

`\color{red}{\text{d)}` `D=x^5-15x^4+16x^3-29x^2+13x`

Ta có: `x=14⇒x-14=0`

`⇒D=x^5-14x^4-x^4+14x^3+2x^3-28x^2-x^2+14x-x`

`⇒D=x^4(x-14)-x^3(x-14)+2x^2(x-14)-x(x-14)-x`

`⇒D=x^4·0-x^3·0+2x^2·0-x·0-x`

`⇒D=0-0+0-0-x`

`⇒D=-x`

`⇒D=-14`

Vậy `D=-14`
~~~~~
`\color{red}{\text{e)}` `E=x^5-100x^4+100x^3-100x^2+100x-9`

Ta có: `x=99⇒x+1=100`

`⇒E=x^5-(x+1)·x^4+(x+1)·x^3-(x+1)·x^2+(x+1)·x-9`

`⇒E=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9`

`⇒E=x-9`

`⇒E=99-9`

`⇒E=90`

Vậy `E=90`