A=\(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

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

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

\(A=\frac{99}{100}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\Rightarrow a=b=c=d\)

\(\Rightarrow M=\frac{a^{2001}\cdot b\cdot c^2}{b^{2004}}=\frac{a^{2004}}{b^{2004}}=1\)

ta có

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)

\(\Rightarrow a=b=c=d\)

.........

chào hen 

Ta có :

\(\frac{1}{\sqrt{1}}>\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{10}\)

.....

\(\frac{1}{\sqrt{99}}>\frac{1}{10}\)

\(\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Cộng vế theo vế ta có :\(\frac{1}{\sqrt{1}}\frac{1}{\sqrt{2}}+......+\frac{1}{\sqrt{99}}+\frac{1}{100}>100.\frac{1}{10}=10\)

Vì \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{cases}}\)  \(\Rightarrow4x\ge0\)

Khi đó: \(\left|x+1\right|=x+1;\left|x+2\right|=x+2;\left|x+3\right|=x+3\)

Thay vào đề bài ta đc:

\(x+1+x+2+x+3=4x\)

\(\Rightarrow3x+6=4x\)

\(\Rightarrow3x-4x=-6\)

\(\Rightarrow-x=-6\)

\(\Rightarrow x=6\)

Vậy \(x=6.\)

đừng giải

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

\(A=5+4+4^2+.....+4^{100}\)

\(A=5+4\left(1+4\right)+4^3\left(1+4\right)+......+4^{99}\left(1+4\right)\)

\(A=5+4\cdot5+4^3\cdot5+......+4^{99}\cdot5\)

\(A=5\left(1+4+4^3+.....+4^{99}\right)⋮5\)

Vậy \(A⋮5\)

\(4A=4\left(1+4+4^2+.........+4^{1000}\right)\)

\(4A=4+4^2+........+4^{1001}\)

\(\Rightarrow4A-A=\left(4+4^2......+4^{1001}\right)-\left(1+4+4^2+......+4^{1000}\right)\)

\(\Rightarrow3A=4^{1001}-1\)

\(\Rightarrow A=\frac{4^{1001}-1}{3}\)

n=1 hoac n=0

n^2=n^5

=>n^2-n^5=0

=>n^2(1-n^5)=0

=>n^2=0 hoặc 1-n^5=0

=>n=0 hoặc n^5=1

=>n=0 hoặc n=1

\(2n+1⋮n+1\\ \Leftrightarrow2\left(n+1\right)-1⋮n+1\\ \Leftrightarrow1⋮n+1\\ \Leftrightarrow n+1\in\left(1\right)=\left\{1;-1\right\}\\ \Leftrightarrow n=0;-2\)

tao có 2n+1 <=>2.(n+1) chia hết cho n+1

suy ra n+1 thuộc Ư(2)={-1;1;-2;2}

ta có bảng 

n+1  -1         1   -2            2

n      -2 loại     0    -3 loại    1

vậy n ={ 0;1}

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{101}{100}=\frac{101}{2}\)

101/2