Cái này là toán lớp 8 chứ không phải lớp 10 nhé

Bài 1: 

1: \(=x^6+27-x^6-9x^4-27x^2-27\)

\(=-9x^4-27x^2\)

2: \(=x^3-9x^2+27x-27-x^3+27+6x^2+12x+6\)

\(=-3x^2+39x+6\)

Bài 2: 

Sửa đề: \(\dfrac{2006^3+1}{2006^2-2005}\)

\(=\dfrac{\left(2006+1\right)\left(2006^2-2006+1\right)}{2006^2-2005}\)

\(=2006+1=2007\)

Ta có:

\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+4+...+100}\)

\(A=3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\right)\)

Đặt \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\), khi đó ta đc:

\(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\)

Vì tổng số hạng bằng (số cuối + số đầu) . số số hạng : 2 nên ta có:

\(B=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+100\right).100:2}\)

\(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\)

\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)

\(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(B=2.\left(1-\frac{1}{101}\right)\)

\(B=2.\frac{100}{101}=\frac{200}{101}\)

Ta có:

\(A=3.B\Rightarrow A=3.\frac{200}{101}=\frac{600}{101}\)

Vậy \(A=\frac{600}{101}\)

để mình giúp  

ĐKXĐ : \(x\ne0;1;-1\)

Ta có :

\(C=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x+1}{x^2-1}\right).\frac{x+2006}{x}\)

\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x+1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2006}{x}\)

\(=\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)

\(=\frac{2x+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)

\(=\frac{\left(x-1\right)^2\left(x+2006\right)}{\left(x-1\right)\left(x+1\right).x}\)

\(=\frac{\left(x-1\right)\left(x+2006\right)}{x\left(x-1\right)}\)

Vậy....

Ta có:

Tập hợp A:
\(A=\left\{1;2;3;5;8\right\}\)

Tập hợp B:

\(B=\left\{-1;0;1;5;9\right\}\)

Mà: \(A\cup B\)

\(\Rightarrow A\cup B=\left\{-1;0;1;2;3;5;8;9\right\}\)

⇒ Chọn B

Chọn B

a) \({4^6}.\sqrt {0,1}  = 1295,2689\)

b) \(\sqrt[8]{{2,{1^{18}} + 1}} - \sqrt {2,{1^{12}} + 1}  =  - 80,4632\)

c) \(\frac{{1,{5^3}}}{{\sqrt[3]{{6,8}}}} = 1,7814\)