a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu

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

=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)

=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

=>\(A=1-\frac{1}{2^{100}}\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\\ \Rightarrow3A=3+3^2+3^3+...+3^{100}+3^{101}\\ \Rightarrow3A-A=3^{101}-1\\ \Rightarrow2A=3^{101}-1\\ \Rightarrow A=\left(3^{101}-1\right).\dfrac{1}{2}\\ \Rightarrow\dfrac{3^{101}}{2}-\dfrac{1}{2}.\)

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)

Ta có: \(3A=3+3^2+3^3+...+3^{99}+3^{100}\)

Khi đó: \(3A-A=3+3^2+3^3+...+3^{99}+3^{100}+3^{101}-\left(1+3+3^2+3^3+...+3^{99}+3^{100}\right)\)

\(=3^{101}-1\)

\(\Leftrightarrow2A=3^{101}-1\)

Vậy \(A=\left(3^{101}-1\right):2\)

em cảm ơn mọi nguoif ạ

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 \(x\in\left\{3;1;5;-1\right\}\)

dễ mà , tk mk trước mk sẽ tk , hứa nà

Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)

=> 2A - A = 1 - \(\frac{1}{2^{100}}\)

<=> A = 1 - \(\frac{1}{2^{100}}\)

\(A=\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\)

\(\Rightarrow2A=1+\frac{1}{2^1}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)

\(A=1-\frac{1}{2^{100}}\)