Let \(\left\{{}\begin{matrix}V_1=4\\V_2=20\end{matrix}\right.\) (km/h)  

We got this :

\(S=V.t\) (with t is the time , S is the length of roads and V is the speed)

\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{S}{V_1}\\t_2=\dfrac{S}{V_2}\end{matrix}\right.\)

From the topic :

\(t_1=t_2+24\)  (minute)

\(\Leftrightarrow\dfrac{S}{V_1}=\dfrac{S}{V_2}+24\)

\(\Leftrightarrow\dfrac{S}{4}=\dfrac{S}{20}+24\)

\(\Leftrightarrow\dfrac{S}{4}=\dfrac{S+24.20}{20}\)

<=> 20S = 4S + 4.20.24

<=> 16S = 4.20.24

<=> S = 120 (km)

So , there are 20 kilometres do you live from the station . 

A = 5 + 5² + 5³ + .................. + 5²⁰¹²

A = (5 + 5² + 5³ + 5⁴) + (5⁵ + 5⁶ + 5⁷ + 5⁸) + ................... + (5²⁰⁰⁹ + 5²⁰¹⁰ + 5²⁰¹¹ + 5²⁰¹²)

A = (5 + 25 + 125 + 625) + 5⁴.(5 + 25 + 125 + 625) + ....................... + 5²⁰⁰⁸.(5 + 25 + 125 + 625)

A = 780.1 + 5⁴.780 + ......................... + 5²⁰⁰⁸.780

A = 780.(1 + 5⁴ + ......................... + 5²⁰⁰⁸)

A = 12.65.(......................)

=> A \(⋮\) 65

1. Assume that \(\sqrt{2}\) is a rational number. That means there exist two integers a and b such that a / b =  \(\sqrt{2}\) 
2. So it can be written as a fractional fraction (a fraction that can not be reduced again): a / b with a, b is two primes together and (a / b)² = 2.
3. From (2) denotes a² / b² = 2 and a² = 2.b².
4. Then a² is even because it is equal to 2.b² (obviously even)
5. From this inference, a must be an even number because a² is the even number of digits (odd numbers are square rooted as odd, even square digits have square root).
6. Since a is even, there are some k satisfactions: a = 2k.
7. Substitution (6) into (3) we have: (2k)² = 2b² <=> 4k² = 2b² <=> 2k² = b².
8. Because 2k² = b² where 2k² is even, then b² is even, this inference b is also even [reason similar to (5)].
9. From (5) and (8) we have: a and b are even numbers, which contradicts the assumption that a / b is the minimal number in (2).
From the contradiction inference:  \(\sqrt{2}\)  is a rational number that is false and must conclude. \(\sqrt{2}\)  is an irrational number.

9 just after 8

7 just before 8 

A = x² - 5x - 2xy + 5y + y² + 4

A = (x² - 2xy + y²) - (5x - 5y) + 4

A = (x - y)² - 5(x - y) + 4

A = 1² - 5.1 + 4

A = 1 - 5 + 4

A = 0 

Apply AM-GM inequality for 2 numbers , we have :

\(\dfrac{a^3}{b}+ab\ge2.\sqrt{\dfrac{a^3}{b}.ab}=2a^2\)

\(\dfrac{b^3}{c}+bc\ge2.\sqrt{\dfrac{b^3}{c}.bc}=2b^2\)

\(\dfrac{c^3}{a}+ca\ge2\sqrt{\dfrac{c^3}{a}.ca}=2c^2\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\)

In another case , we have this :

\(a^2+b^2+c^2\ge ab+bc+ca\) (The consequence of AM-GM)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2ab+2bc+2ca\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

We have :

\(x^2-2x+5=x^2-2x+1+4\)

                      \(=\left(x-1\right)^2+4\ge4\)

\(\Rightarrow H=\dfrac{1}{x^2-2x+5}\le\dfrac{1}{4}\)

\(\Rightarrow Max_H=\dfrac{1}{4}\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

We have :

\(\left(x-1\right)^4=\left(x-1\right)^6\)

\(\Rightarrow\left(x-1\right)^6-\left(x-1\right)^4=0\)

\(\left(x-1\right)^4.\left[\left(x-1\right)^2-1\right]=0\)

So \(\left(x-1\right)^4=0\)           or          \(\left(x-1\right)^2-1=0\)

\(\Leftrightarrow x-1=0\)                or           \(\left(x-1\right)^2=1\)

\(\Leftrightarrow x=1\)                       or            \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

So , x = {0,1,2}

A = (x² + y²)³ + (z² - x²)³ - (y² + z²)³

A = [(x² + y²) + (z² - x)²]³ + 3.(x² + y²).(z² - x²).(x² + y² + z² - x²) - (y² + z²)³

A = (y² + z²)³ + 3.(x² + y²).(x + z)(z - x).(y² + z²) - (y² + z²)³

A = 3.(x² + y²).(x + z)(z - x).(y² + z²) 

We have :

M = 2(a³ + b³) - 3(a² + b²)

M = 2.(a + b).(a² - ab + b²) - 3.[(a + b)² - 2ab]

M = 2(a² - ab + b²) - 3.1 + 3.2ab

M = 2a² - 2ab + 2b² - 3 + 6ab

M = 2a² + 4ab + 2b² - 3

M = 2.(a² + 2ab + b²) - 3

M = 2.(a + b)² - 3

M = 2 - 3 = -1 

We have :

a³ + b³ + c³ = 3abc 

a³ + b³ + c³ - 3abc = 0

(a³ + 3a²b + 3ab² + b³) + c³ - 3a²b - 3ab² - 3abc = 0

(a + b)³ + c³ - 3ab.(a + b + c) = 0

[(a + b) + c].[(a + b)² - c.(a + b) + c²] - 3ab(a + b + c) = 0

(a + b + c).[a² + 2ab + b² - ac - bc + c²] - 3ab(a + b + c) = 0

(a + b + c).[a² + b² + c² - bc - ca + 2ab - 3ab] = 0

(a + b + c).[a² + b² + c² - ab - bc - ca] = 0

That is right , because a + b + c = 0

So a³ + b³ + c³ = 3abc

With a + b + c = 0

Ok , Do it , I support :v 

\(S=1-2+3-4+....+2009-2010\)

Denote :

\(S'=1+3+.....+2009\) ; \(S"=-2-4-....-2010=-\left(2+4+....+2010\right)\)

We have :

\(S'=\dfrac{\left[\left(2009-1\right):2+1\right].\left(2009+1\right)}{2}=\dfrac{1005.2010}{2}=1005^2\)

\(S"=-\dfrac{\left[\left(2010-2\right):2+1\right].\left(2010+2\right)}{2}=-\dfrac{1005.2012}{2}=-1005.1006\)

So :

\(S=S'+S"=1005^2-1005.1006=1005.\left(1005-1006\right)=-1005\)

a) \(A=5+5^2+5^3+....+5^{100}\)

\(5A=5^2+5^3+....+5^{100}+5^{101}\)

\(5A-A=\left(5^2+5^3+5^4+......+5^{101}\right)-\left(5+5^2+5^3+....+5^{100}\right)\)

\(4A=5^{101}-5\)

\(A=\dfrac{5^{101}-5}{4}\)

b) \(A=5+5^2+5^3+....+5^{100}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\)

\(A=\left(5+25\right)+5^2.\left(5+25\right)+....+5^{98}.\left(5+25\right)\)

\(A=30+5^2.30+.....+5^{98}.30\)

\(A=30.\left(1+5^2+....+5^{98}\right)⋮30\)

\(\Rightarrow A⋮30\)

Phan Thanh Tinh :3 i know that , so , what are you going to do :v , "WHAT ARE YOU GOING TO DO ?"

Phan Thanh Tinh : may be from that :v 

:v , This web is just starts in few months , so you should not demanding too much ~ 

I am agree :v 

Let \(A=\dfrac{a^3+b^3+c^3+d^3}{a+b+c+d}\)

\(A=\dfrac{a^3}{a+b+c+d}+\dfrac{b^3}{a+b+c+d}+\dfrac{c^3}{a+b+c+d}+\dfrac{d^3}{a+b+c+d}\)

\(A=\dfrac{a^4}{a^2+ab+ac+ad}+\dfrac{b^4}{ab+b^2+bc+bd}+\dfrac{c^4}{ac+bc+c^2+cd}+\dfrac{d^4}{ad+bd+cd+d^2}\)

Following Schwarz's inequality , we have :

\(A\ge\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)}=\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{\left(a+b+c+d\right)^2}\)

Following Bunyakovsky's inequality , we have :

\(\left(a^2+b^2+c^2+d^2\right)\left(1+1+1+1\right)\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow4\left(a^2+b^2+c^2+d^2\right)\ge\left(a+b+c+d\right)^2\)

\(\Rightarrow A\ge\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{4\left(a^2+b^2+c^2+d^2\right)}=\dfrac{a^2+b^2+c^2+d^2}{4}\ge\dfrac{4\sqrt[4]{a^2b^2c^2d^2}}{4}=\dfrac{4.1}{4}=1\)

\(\Rightarrow a^3+b^3+c^3+d^3\ge a+b+c+d\)

Equality occurs when :

a = b = c = d = 1

10x(x - y) + 8y(y - x)

= 10x² - 10xy + 8y² - 8xy

= 10x² - 18xy + 8y²

= 2.(5x² - 9xy + 4y²)

= 2.(4x² + x² - 2.2x.2y - xy + 4y²)

= 2.[(4x² - 2.2x.2y + 4y²) + x² - xy]

= 2.[ (2x - 2y)²  + x(x - y)]

= 2.[ 4.(x - y)² + x(x - y)]

= 2.(x - y).[4.(x - y) + x]

= 2.(x - y).[5x - y]