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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/ \(x=99\Rightarrow100=x+1\)
\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)
\(=x-9=99-9=90\)
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
a,
\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)
Thay $x=\dfrac12$ vào $A$, ta được:
\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)
Vậy $A=\dfrac94$ khi $x=\dfrac12$.
b,
\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)
Thay $x=1$ vào $B$, ta được:
\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)
Vậy $B=0$ khi $x=1$.
$Toru$
Ta có x = 99
=> x + 1 = 100
Khi đó A = x5 - 100x4 + 100x3 - 100x2 + 100x - 9
= 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
Thay x = 99 vào A
=> A = x - 9 = 99 - 9 = 90
Vậy A = 90
Ta có : \(x=99\Rightarrow100=x+1\)
\(A=x^5-100x^4+100x^3-100x^2+100x-9\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)
\(=x-9\)hay \(99-9=90\)
Vậy \(A=90\)
x = 99 suy ra 100 = x +1
A= x^5 - (x + 1)x^4 + (x + 1)x^3 - (x+1)x^2 + (x +1)x - 9
A= x^5 - x^5 - x^4 + x^4 +x^3 - x^3 -x^2 +x^2 + x - 9
A= x - 9 = 99 - 9 = 90
\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)
\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)
a)\(A=x^5-100x^4+100x^3-100x^2+100x-9\)
\(A=x^5-(99+1)x^4 +(99+1)x^3-(99+1)x^2+(99+1)x-9\)
Tại x=99 , ta có :
\(A=x^5 - (x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-9\)
\(A=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)
\(A=x-9\)
Thay x = 99 vào biểu thức A ta có :
\(A=99-9=90\)
a, \(A=x^5-100x^4+100x^3-100x^2+100x-9\)
\(=x^5-99x^4-x^4+99x^3+x^3-99x^2-x^2+99x+x-9\)
\(=x^4\left(x-99\right)-x^3\left(x-99\right)+x^2\left(x-99\right)-x\left(x-99\right)+x-9\)\(=\left(x^4-x^3+x^2-x\right)\left(x-99\right)+x-9\)
Thay x = 99
\(\Rightarrow A=90\)
Vậy A = 90 tại x = 99
b, \(B=x^7-26x^6+27x^5-47x^4-77x^3+50x^3+50x^2+x-24\)
\(=x^7-25x^6-x^6+25x^5+2x^5-50x^4+3x^4-75x^3-2x^3+50x^2+x-24\)
\(=x^6\left(x-25\right)-x^5\left(x-25\right)+2x^4\left(x-25\right)+3x^3\left(x-25\right)-2x^2\left(x-25\right)+x-24\)
\(=\left(x^6-x^5+2x^4+3x^3-2x^2\right)\left(x-25\right)+x-24\)
Thay x = 25
\(\Rightarrow B=1\)
Vậy B = 1 tại x = 25
