\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

1,

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(1A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(A=2^{32}-1\)

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)

\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Đặt : \(P=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

3(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^{16}-1\right)\left(2^{16}+1\right)\)

=\(2^{32}-1\)

Câu 21: So sánh M = 2³² và N = (2 + 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)

A. M > N                      B. M < N                    C. M = N                         D. M = N – 1

Câu 22: Tìm giá trị lớn nhất của biểu thức B = 4 – 16x² – 8x

A. 5                         B. -5                               C. 8                                       D.-8  

Câu 23: Biểu thức E = x² – 20x +101 đạt giá trị nhỏ nhất khi

A. x = 9                           B. x = 10                 C. x = 11                              D.x = 12

Câu 24: Kết quả của phép chia 15x³y⁴ : 5x²y² là

A. 3xy²                            B. -3x²y                        C. 5xy                                  D. 15xy²

Câu 25: Kết quả của phép chia (6xy² + 4x²y – 2x³) : 2x là

A. 3y² + 2xy – x²                B. 3y² + 2xy + x²           C. 3y² – 2xy – x²                        D. 3y² + 2xy

các bn giúp mình nhé

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

Ta có

N   =   ( 2   +   1 ) ( 2 2   +   1 ) ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     ( 2 16   +   1 )   =   3 ( 2 2   +   1 ) ( 2 4   +   1 ) ( 2 8   +   1 )     ( 2 16   +   1 )   =   [ ( 2 2   –   1 ) ( 2 2   +   1 ) ] ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 4   –   1 ) ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 8   –   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     =   ( 2 16   -   1 ) ( 2 16   +   1 )   = 2 16 2 − 1 = 2 32 − 1 M à   2 32 − 1 > 2 32 ⇒   N < M

Đáp án cần chọn là: A

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!