\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}=1-\dfrac{1}{1024}=\dfrac{1023}{1024}\)

We have:

\(B\cdot101=\dfrac{101}{1\cdot102}+\dfrac{101}{2\cdot103}+...+\dfrac{101}{101\cdot400}\)

\(=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...+\dfrac{1}{299}-\dfrac{1}{400}\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+\dfrac{1}{104}+...+\dfrac{1}{400}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\)

\(=\left(1-\dfrac{1}{300}\right)+\left(\dfrac{1}{2}-\dfrac{1}{301}\right)+...+\left(\dfrac{1}{101}-\dfrac{1}{400}\right)\)

\(=\dfrac{299}{1\cdot300}+\dfrac{299}{2\cdot301}+...+\dfrac{299}{101\cdot400}\)

\(=299\left(\dfrac{1}{1\cdot300}+\dfrac{1}{2\cdot301}+...+\dfrac{1}{101\cdot400}\right)\)

So B.101 = A.299

=> 101 = A/B . 299

=> A/B = 101/299

So \(\dfrac{A}{B}=\dfrac{101}{299}\)

\(\dfrac{121212}{424242}=\dfrac{121212\div60606}{424242\div60606}=\dfrac{2}{7}\)

\(\dfrac{187187187}{221221221}=\dfrac{187187187\div170117017}{221221221\div170117017}=\dfrac{11}{13}\)

\(\dfrac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050+70707}=\dfrac{393939-10101}{505050+70707}=\dfrac{10101\left(39-1\right)}{10101\left(50+7\right)}=\dfrac{39-1}{50+7}=\dfrac{38}{57}=\dfrac{2}{3}\)

We have:

a⁷ = b⁸  [1]

\(\Rightarrow b=\dfrac{a^7}{b^7}=\left(\dfrac{a}{b}\right)^7\)

Because b is a natural number,\(a⋮b\), put a = bk \(\left(k\in N\right)\)

Because b > 1 => \(\dfrac{a}{b}>1\Rightarrow k\ge2\) [2]

Replace a = bk in [1] 

=> b⁷k⁷ = b⁸ => k⁷ = b     [3]

From [2] and [3] 

=> \(b\ge2^7\)

But the question is that the smallest number should be b = 2⁷ = 128

=> a = bk = 128. 2 = 256

So a = 256 and b = 128 

Are you OK, Lê Quốc Trần Anh?

Thanks Phan Thanh Tinh, I will revised to \(\dfrac{13}{10^7-8}>\dfrac{13}{10^8-7}\Rightarrow\dfrac{10^7+5}{10^7-8}>\dfrac{10^8+6}{10^8-7}\)

We have:

\(\dfrac{10^7+5}{10^7-8}=1+\dfrac{13}{10^7-8}\)

\(\dfrac{10^8+6}{10^8-7}=1+\dfrac{13}{10^8-7}\)

Because 10⁷ - 8 < 10⁸ - 7

=> \(\dfrac{13}{10^7-8}< \dfrac{13}{10^8-7}\Rightarrow\dfrac{10^7+5}{10^7-8}< \dfrac{10^8+6}{10^8-7}\)

181 - {-85 + [220 + (-65-35)]}

= 181 - {-85 + [220 + (-100)]}

= 181 - {-85 + 120}

= 181 - 35

= 146

+ The smallest number divisible by 4: 1023457896

+ The biggest number divisible by 4: 9876543120

Because the numbers divisible by 4 will have two digits end that numbers so we can write this numbers faster. 

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1000^{10}=10^{30}\)

\(2^{100}=2^{31}.2^6.2^{63}=2^{31}.64.512^7< 2^{31}\cdot125\cdot625^7=2^{31}\cdot5^3\cdot\left(5^4\right)^7=2^{31}\cdot5^{31}=10^{31}\)

So \(10^{30}< 2^{100}< 10^{31}\)

<=> 2¹⁰⁰ is a number that has 31 digits

\(\dfrac{4}{11}< \dfrac{x}{20}< \dfrac{5}{11}\Leftrightarrow\dfrac{80}{220}< \dfrac{11x}{220}< \dfrac{100}{220}\)

=> 80 < 11x < 100

=> \(11x\in\left\{81,82,83,...,99\right\}\)

\(\Rightarrow x\in\left\{\dfrac{81}{11};\dfrac{82}{11};\dfrac{83}{11};...;8;\dfrac{89}{11};...;9\right\}\)

If x is a integer, x = 8 or x = 9

In the isosceles triangle will have 2 corners are equal so we have:

eg. \(\Delta ABC\) and corner ABC = corner ACB, we have:

\(\widehat{ABC}=\widehat{ACB}=\dfrac{180^o-\widehat{BAC}}{2}\)

Butthe corner BAC > 0^o

\(\Rightarrow\dfrac{180^o-\widehat{BAC}}{2}< \dfrac{180^o}{2}=90^o\)

So in isosceles triangle , bases angles less 90^o

27.8.5³.64 = 3³.2³.5³.4³ = \(\left(2\cdot3\cdot4\cdot5\right)^3=120^3\)

We have a,b,c,d are four digits so \(0\le a,b,c,d\le9\)

Because a^a + [b,c,d] = 100

=> a^a \(\le100\)=> a \(\in\left\{1,2,3\right\}\)

If a = 1

=> [b,c,d] = 100 - 1 = 99 = 9.11

But 11 has two digits so b,c,d unsatisfactory 

If a = 2

=> [b,c,d] = 100 - 2² = 96 = 3.32

But 32 has two digits so b,c,d unsatisfactory 

If a = 3

=> [b,c,d] = 100 - 3³ = 73 

But 73 has two digits so b,c,d unsatisfactory 

Therefore, there are no numbers satisfying the question 

A = 1+2+3+...+1000

= \(\dfrac{1000}{2}\cdot\left(1+1000\right)=500\cdot1001=500500\)have 6 digits

And B \(=1\cdot2\cdot3\cdot...\cdot11=\left(1\cdot2\cdot3\cdot4\cdot5\cdot6\right)\cdot7\cdot8\cdot...\cdot11\)

We have: 1.2.3.4.5.6 have 3 digits

And B will have:

3 + 1 + 1 + 1 + 1 + 1 = 8 [digits]

So A < B

1.2.3.4....20 = \(\left(1.2.3.4.5.6\right).7.8.9.10...20\)

We have: 1.2.3.4.5.6 is a number has three digits

After that, when we multiplied by a next number, we will have a number has more 1 digits than the number before. 

=> \(\left(1.2.3.4.5.6\right).7.8.9....20\)have 19 digits

So A have 19 digits

And B = 1+2+3+...+1000000

= 1000000/2 * \(\left(1+1000000\right)\)= 500 000 * 1000001 = 500000500000 have 12 digits

=> A > B 

Because ABCD is a rectangle so AD = BC = 5cm

The total area of the two triangles AMD and BNC are:

\(\dfrac{5\cdot2}{2}+\dfrac{5\cdot2}{2}=5+5=10\left(cm^2\right)\)

The area of the MNCD quadrilateral is:

\(60-10=50\left(cm^2\right)\)

Answer: 50cm²

\(\dfrac{3-x}{100}-1=\dfrac{2-x}{101}+\dfrac{1-x}{102}\)

\(\Leftrightarrow\dfrac{3-x}{100}-\dfrac{100}{100}=\dfrac{102\left(2-x\right)}{101\cdot102}+\dfrac{101\left(1-x\right)}{101\cdot102}\)

\(\Rightarrow\dfrac{3-x-100}{100}=\dfrac{204-102x+101-101x}{10302}\)

\(\Rightarrow\dfrac{3-x-100}{100}=\dfrac{204+101-102x-101x}{10302}\)

\(\Rightarrow\dfrac{-97-x}{100}=\dfrac{305-203x}{10302}\)

=> 10302[-97 - x] = 100[305 - 203x]

=> -999294 - 10302x = 30500 - 20300x

=> -999294 - 30500 = - 20300x + 10302x

=> -1029794 = -9998x

=> x = 103

So x = 103

Because this topic is not to say that the faces should be different so that their sum is the smallest, then the remaining 3 [can be said to be x, y, z] smallest and satisfy:
  - 12*x = 9*y = 6*z
=> x = y = z = 0 is the smallest
So the sum is: 0 + 0 + 0 + 6 + 9 + 12 = 27
Answer: 27

a,

Put d is common divisor of 2n + 1 and 4n+3, we have:

\(\left\{{}\begin{matrix}2n+1⋮d\\4n+3⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\left(2n+1\right)⋮d\\4n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4n+2⋮d\\4n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left(4n+3\right)-\left(4n+2\right)=1⋮d\)

=> d = \(\pm1\)

b,

Put d is common divisor of 4n+1 and 12n+7, we have:

\(\left\{{}\begin{matrix}4n+1⋮d\\12n+7⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3\left(4n+1\right)⋮d\\12n+7⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12n+3⋮d\\12n+7⋮d\end{matrix}\right.\)

\(\Rightarrow\left(12n+7\right)-\left(12n+3\right)=4⋮d\)

But 12n+7 and 12n+3 is odd so d is odd too.

=> d = \(\pm1\)

c,

Put d is common divisor of 7n+4 and 9n+5, we have:

\(\left\{{}\begin{matrix}7n+4⋮d\\9n+5⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}9\left(7n+4\right)⋮d\\7\left(9n+5\right)⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}63n+36⋮d\\63n+35⋮d\end{matrix}\right.\)

\(\Rightarrow\left(63n+36\right)-\left(63n+35\right)=1⋮d\)

\(\Rightarrow d=\pm1\)

True or False? 

From page 1 to page 9 have 9 digits

From page 10 to page 99 have:

\(\left(99-10+1\right)\times2=180\) [digits]

From page 100 to page 120 have:

\(\left(120-100+1\right)\times3=63\) [digits]

To number all pages that need the number of digits are:

9 + 180 + 63 = 252 [digits]

Answer: 252 digits