Ta có:  

\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right).....+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)....\left(2^{256}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)....\left(2^{256}+1\right)\)

\(......\)

\(=\left[\left(2^{256}\right)^2-1\right]+1\)

\(=2^{512}\)

A=2⁵¹²

A=(2-1)(2+1)*...*(2^256+1)+1

=(2^2-1)(2^2+1)*...*(2^256+1)+1

=(2^4-1)(2^4+1)*...*(2^256+1)+1

=(2^8-1)(2^8+1)(2^16+1)(2^32+1)*....*(2^256+1)+1

=(2^16-1)(2^16+1)*....*(2^256+1)+1

=(2^32-1)(2^32+1)*...*(2^256+1)+1

=(2^64-1)(2^64+1)(2^128+1)(2^256+1)+1

=(2^128-1)(2^128+1)(2^256+1)+1

=(2^256-1)(2^256+1)+1

=2^512

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

đặt A = (2 + 1)(2² + 1)...(2²⁵⁶ + 1).

khi đó (2 - 1)A = (2 -1)(2 + 1)(2² + 1)...(2²⁵⁶ + 1)

suy ra A = 2²⁵⁷ - 1 (dùng hiệu hai bình phương).

nên biểu thức đã cho là A + 1 = 2²⁵⁷.

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

A=(ghi lại biieur thức)

2A=2+1+1/2+1/2^2+….+1/2^2011

2A-A=A=(2+1+1/2+1/2^2+….+1/2^2011)-(1+1/2+1/2^2+...+1/2^2012)

A=2-1/2^2012

1/2 A= 1/2+1/2^2+1/2^3+1/2^4+...........+1/2^2013

=>A-1/2A= 1 -1/2^2013

=>1/2A=1 -1/2^2013

=>A=(1 - 1/2^2013) : 1/2

I don't know .

\(=\left(a^2-1\right)^3-\left(a^6-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)