`@` `\text {Ans}`
`\downarrow`
`b,`
\(B=x^6 - 20x^5 - 20x^4 - 20x^3 - 20x^2 - 20x + 3\) tại `x=21`
Ta có: `20 = 21 - 1 => 20 = x-1`
Thay `20 = x-1` vào, ta có:
\(x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)
`=`\(x^6-x^6+x^5-x^5+x^4-x^4+...+x+3\)
`=`\(x+3\)
`=`\(21+3=24\)
Vậy, `B=24`
`c,`
`C=`\(x^7-26x^6+27x^5-47x^4-77x^3+50x^2+x-24\) tại `x=25`
`=`\(x^7-25x^6-x^6+25x^5+2x^5-50x^4+3x^4-75x^3-2x^3+50x^2+x-24\)
`=`\(x^6\left(x-25\right)-x^5\left(x-25\right)+2x^4\left(x-25\right)+3x^3\left(x-25\right)-2x^2\left(x-25\right)+x-24\)
`=`\(\left(x^6-x^5+2x^4+3x^3-2x^2\right)\left(x-25\right)+x-24\)
Thay `x=25` vào bt C, ta được:
\(\left(25^6-25^5+2\cdot25^4+3\cdot25^3-2\cdot25^2\right)\left(25-25\right)+25-24\)
`=`\(\left(25^6-25^5+2\cdot25^4+3\cdot25^3-2\cdot25^2\right)\cdot0+1\)
`= 0+1=1`
Vậy, `C=1.`
