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ta có: 20=21-1=x-1
B=x6-20x5-20x4-20x3-20x2-20x+3
= x6-(x-1)x5-(x-1)x4-(x-1)x3-(x-1)x2-(x-1)x+3
=x6-x6+x5-x5+x4-x4+x3-x3+x2-x2+x+3
=x+3
=21+3
=24
a) Có x = 99 => x+1 = 100
A = x5 - (x+1)x4 + (x+1)x3 + (x+1)x2 + (x+1)x - 9
= x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - 9
= x - 9
=> A = 90
b) Chữa đề: x6 - 20x5 - 20x4 - 20x3 - 20x2 - 20x + 3
Có: x = 21 => x-1 = 20
B = x6 - (x-1)x5 - (x-1)x4 - (x-1)x3 - (x-1)x2 - (x-1)x + 3
= x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x + 3
= x + 3
=> B = 24
`#3107.101107`
`5x^2y(3x^3 - 4y + 5xy) - 15x^5y + 20x^2y^2`
`= 15x^5y - 20x^2y^2 + 25x^3y^2 - 15x^5y + 20x^2y^2`
`= (15x^5y - 15x^5y) + (-20x^2y^2 + 20x^2y^2) + 25x^3y^2`
`= 25x^3y^2`
_______
`(8x^5y^2 + 4x^3y^3 - 2x^6y^2) \div 2x^3y`
`= 4x^2y + 2y^2 - x^3y`
a/ \(x=99\Rightarrow100=x+1\)
\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)
\(=x-9=99-9=90\)
b/ Tương tự \(20=x-1\)
\(B=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)
\(=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+3\)
\(=x+3=24\)
c/ \(26=x+1;27=x+2;47=2x-3;77=3x+2;50=2x\)
\(C=x^7-\left(x+1\right)x^6+\left(x+2\right)x^5-\left(2x-3\right)x^4-\left(3x+2\right)x^3+2x.x^2+x-24\)
\(=x-24=1\)
a/ x=99⇒100=x+1x=99⇒100=x+1
A=x5−(x+1)x4+(x+1)x3−(x+1)x2+(x+1)x−9A=x5−(x+1)x4+(x+1)x3−(x+1)x2+(x+1)x−9
=x5−x5−x4+x4+x3−x3−x2+x2+x−9=x5−x5−x4+x4+x3−x3−x2+x2+x−9
=x−9=99−9=90=x−9=99−9=90
b/ Tương tự 20=x−120=x−1
B=x6−(x−1)x5−(x−1)x4−(x−1)x3−(x−1)x2−(x−1)x+3B=x6−(x−1)x5−(x−1)x4−(x−1)x3−(x−1)x2−(x−1)x+3
=x6−x6+x5−x5+x4−x4+x3−x3+x2−x2+x+3=x6−x6+x5−x5+x4−x4+x3−x3+x2−x2+x+3
=x+3=24=x+3=24
c/ 26=x+1;27=x+2;47=2x−3;77=3x+2;50=2x26=x+1;27=x+2;47=2x−3;77=3x+2;50=2x
C=x7−(x+1)x6+(x+2)x5−(2x−3)x4−(3x+2)x3+2x.x2+x−24C=x7−(x+1)x6+(x+2)x5−(2x−3)x4−(3x+2)x3+2x.x2+x−24
=x−24=1=x−24=1
Sửa đề: x=19
x=19 nên x+1=20
\(A=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2018\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2018\)
=x-2018
=19-2018=-1999