x =99 => 100 = x + 1 thay vào ta có 

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

= x - 9

= 99 -9 

= 90

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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

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x=99=>x+1=100

A=x⁵-(x+1)x⁴+(x+1)x³-(x+1)x²+(x+1)x-9

A=x⁵-x⁵-x⁴+x⁴+x³-x³-x²+x²+x-9

A=99-9

A=90

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 \(A=90\)

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

Ta có: x=99

⇒x+1=100

Thay vào biểu thức ta có:

B=\(\text{x}^{\text{5}}\)- (x+1).\(\text{x}^{\text{4}} + ( x+1). \text{x}^{\text{3}} - (x+1). \text{x}^{\text{2}}\)+ (x+1).x - 9

B=\(\text{x}^{\text{5}} - \text{x}^{\text{5}} - \text{x}^{\text{4}} + \text{x}^{\text{4}} + \text{x}^{\text{3}} - \text{x}^{\text{3}} - \text{x}^{\text{2}} + \text{x}^{\text{2}} + x -9\)

B= x-9

vậy giá trị của biểu thức tại x=99 là

B= 99-9=90

Dễ thấy 100=99+1=x+1,thay vào biểu thức ta có:

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=99-9=90

Vậy biểu thức = 90 tại x=99

Đề bài phải là tính chứ?x có sẵn tìm kiểu j

Cậu ơi tìm x mà còn cho x = 99

=> x = 99