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.

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

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

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

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

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 \(x\in\left\{3;1;5;-1\right\}\)

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 .

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

\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)

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

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

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

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

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

b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)

=>\(x^2+x+1-x⋮x^2+x+1\)

=>\(x⋮x^2+x+1\)

=>\(x^2+x⋮x^2+x+1\)

=>\(x^2+x+1-1⋮x^2+x+1\)

=>\(-1⋮x^2+x+1\)

=>\(x^2+x+1\in\left\{1;-1\right\}\)

=>\(x^2+x+1=1\)

=>x²+x=0

=>x(x+1)=0

=>\(x\in\left\{0;-1\right\}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

2A = 2 + 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹¹

mà A = 1 + 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰¹²

2A - A = 2 - 1/2²⁰¹²

A = 2 - 1/2²⁰¹²

Ta có A=1+1/2+1/2^2+1/2^3+........+1/2^2012

=>2A=2+1+1/2+1/2^2+.......+1/2^2011

=>2A-A=(2+1+1/2+1/2^2+.....+1/2^2011)-(1+1/2+1+1/2^2+1/2^3+.....+1/2^2012)

=>A=\(2-\frac{1}{2^{2012}}\)

\(A=\frac{2^{2013-1}}{2^{2012}}\)

2A=2+1+1/2+1/2^2+1/2^3+...+1/2^2011
2A-A=(2+1+1/2+1/2^2+1/2^3+...+1/2^2011)-(1+1/2+1/2^2+1/2^3+...+1/2^2012)
A=2-2/2012
k cho mik nhé

1+1/2012