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

x=21

=>x-1=20

B=x^6-x^5(x-1)-x^4(x-1)-...-x(x-1)+3

=x^6-x^6+x^5-x^5+x^5-...-x^2+x+3

=x+3

=21+3=24

P=x3+y3+2xy=(x+y)3−3xy(x+y)+2xy=2013−601xy" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">P=x3+y3+2xy=(x+y)3−3xy(x+y)+2xy=2013−601xy

S=xy=x(201−x)" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=xy=x(201−x)

1≤x≤200" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">1≤x≤200

S=200−(x−1)(x−200)≥0⇒Smin=200" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=200−(x−1)(x−200)≥0⇒Smin=200

x≤y⇒x≤100" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x≤y⇒x≤100

S=100.101−(x−100)(x−101)≤100.101⇒Smax=100.101" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=100.101−(x−100)(x−101)≤100.101⇒Smax=100.101

20x2+20x3+20x4+20 

= 200

20 x 2 + 20 x 3 + 20 x 4 +20

=40 + 60 + 80 + 20

=100+80+20

=180+20

=200

Trả lời :

200

Hokt~

:)

a) 20x4+20x3+20x3

= 20 x ( 4+3 + 3)

= 20 x10

= 200

 b) (6x9-54)x(25 +26+27+.........41+42)

= 0 x (25 +26+27+.........41+42)

= 0 

20x4+20x3+20x3

= 20 x ( 4+3 + 3)

= 20 x10

= 200

 b) (6x9-54)x(25 +26+27+.........41+42)

= 0 x (25 +26+27+.........41+42)

= 0 

a) = 20 x ( 4 + 3 + 3 )

   = 20 x 10

  = 200

b) = 0 x ( 25+26+27+....41+42)

= x

(x:10)x5=1000-80

(x:10)x5=920

(x:10)    =920 : 5

(x:10)    =184

    x        =184 x 10

    x        =1840

(x:10)x5=1000-20x4

(x:10)x5=920

(x:10)=920:5

(x:10)=184

x        =184x10

x         =1840

(x:10)x5 =1000-20x4 

(x:10)x5=1000-80

(x:10)x5=920

(x:10)    =920:5  

x:10       =184 

x           =184 x10

x            =1840

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

a) Vì\(x=99\Rightarrow x+1=100\)

Thay x+1=100 vào biểu thức A ta được :

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+9\)

\(=x+9\)

\(=99+9\)

\(=108\)

b) Tương tự

\(A=x^5-100x^4+100x^3-100x^2+100x-9\)

\(\Rightarrow A=x^5-99x^4-x^4+99x^3+x^3-99x^2-x^2+99x+x-9\)

\(\Rightarrow A=x^4\left(x-99\right)-x^3\left(x-99\right)+x^2\left(x-99\right)+x\left(x-99\right)-9\)

\(\Rightarrow A=x^4\left(99-99\right)-x^3\left(99-99\right)+x^2\left(99-99\right)+x\left(99-99\right)-9\)

\(\Rightarrow A=x^4.0-x^3.0+x^2.0+x.0-9\)

\(\Rightarrow A=0-0+0+01-9=-9\)

Theo cách của Lê Tài Bảo Châu thì, câu b)

\(x=21\Rightarrow x-1=20\)

\(\Rightarrow B=x^6-20x^5-20x^4-20x^3-20x^2-20x+3\)

\(=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)

\(=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+3\)

= x + 3 = 21 + 3 = 24