Suppose \(\dfrac{x}{x-4}\Rightarrow a=\dfrac{x}{\overline{x-4|x}}\)

a, 

a = \(-\dfrac{3}{13}=\dfrac{-3}{13}=\dfrac{3}{-13}\)

\(\Rightarrow\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Easy see x = 3 => a \(=-\dfrac{3}{13}\)

So x = 3

b,

a = 1/6 

=> a - 4 \(\ge0\)

=> a > 4

\(\Rightarrow5\le x\le9\)

\(\Rightarrow1\le x-4\le5\)

\(\Rightarrow a\in\left\{\dfrac{5}{15};\dfrac{6}{26};\dfrac{7}{37};\dfrac{8}{48};\dfrac{9}{59}\right\}\)

Have 8/48 = 1/6

So a = 8/48

=> x = 8

The number of time ahead of Canadaxe does Vicki finish is:

29'46'' - 28'47'' = 59''

Answer: 59 seconds

a \(⋮4\)

=> a \(⋮\)divisor of 4

=> a \(⋮\pm1;\pm2;\pm4\)

So a \(⋮4\Rightarrow a⋮\pm1;\pm2;\pm4\)

1^st way:

10 + 20 + 30 + 40 + 50 + 60

= [10 + 60] + [20 + 50] + [30 + 40]

= 70 + 70 + 70 

= 70.3 

= 210

2^nd way:

10 + 20 + 30 + 40 + 50 + 60

= \(\left[10\cdot\left(1+2+3+4+5+6\right)\right]\)

\(=10\cdot21\)

= 210

Because all number is equal

=> Their average equals one of them

IN there is 10

Their average are:

\(\dfrac{6+8+16}{3}=\dfrac{30}{3}=10\)

M must \(=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\)

Because abc = 1

=> exist 3 numbers x,y,z so that: \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\)

\(\Rightarrow M=\dfrac{\dfrac{x}{y}}{\dfrac{x}{y}\cdot\dfrac{y}{z}+\dfrac{x}{y}+1}+\dfrac{\dfrac{y}{z}}{\dfrac{y}{z}\cdot\dfrac{z}{x}+\dfrac{y}{z}+1}+\dfrac{\dfrac{z}{x}}{\dfrac{z}{x}\cdot\dfrac{x}{y}+\dfrac{z}{x}+1}\)

\(\Rightarrow M=\dfrac{xz}{xy+yz+zx}+\dfrac{xy}{yz+zx+xy}+\dfrac{yz}{zx+xy+yz}\)

\(=\dfrac{xz+xy+yz}{xy+yz+zx}=1\)

So if abc = 1

=> \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}=1\)

We have:

\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}=\dfrac{ab}{2c}+\dfrac{bc}{2a}+\dfrac{ca}{2b}+\dfrac{ca}{2b}+\dfrac{ab}{2c}+\dfrac{bc}{2a}\)

\(\ge2\sqrt{\dfrac{ab}{2c}\cdot\dfrac{ca}{2b}}+2\sqrt{\dfrac{bc}{2a}\cdot\dfrac{ca}{2b}}+2\sqrt{\dfrac{ab}{2c}\cdot\dfrac{bc}{2a}}=a+b+c\)

\(\Rightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)

The "=" occurs when a = b = c

1^st way:

100 : 5 + 50 : 5

= [100 + 50] : 5

= 150 : 5

= 30

2^nd way:

100 : 5 + 50 : 5

= 20 + 10

= 30

Alternate plus 1 to each number of the left and the right of the inequation, we have:

a, The left: 

\(\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1\)

\(=\dfrac{x+5+2006}{2006}+\dfrac{x+4+2007}{2007}+\dfrac{x+3+2008}{2008}\)

\(=\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}\)

\(=\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

So the left plus 3 is: \(\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

* The right:

\(\dfrac{x+9}{2002}+1+\dfrac{x+1}{2010}+1+\dfrac{x+11}{2000}+1\)

\(=\dfrac{x+9+2002}{2002}+\dfrac{x+1+2010}{2010}+\dfrac{x+11+2000}{2000}\)

\(=\dfrac{x+2011}{2002}+\dfrac{x+2011}{2010}+\dfrac{x+2011}{2000}\)

\(=\left(x+2001\right)\left(\dfrac{1}{2002}+\dfrac{1}{2010}+\dfrac{1}{2000}\right)\)

So the right plus 3 is: \(\left(x+2001\right)\left(\dfrac{1}{2002}+\dfrac{1}{2010}+\dfrac{1}{2000}\right)\)

But \(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}< \dfrac{1}{2002}+\dfrac{1}{2010}+\dfrac{1}{2000}\)

because \(\dfrac{1}{2006}< \dfrac{1}{2002};\dfrac{1}{2007}< \dfrac{1}{2000}and\dfrac{1}{2008}>\dfrac{1}{2010}\)

have two numbers of the left is smaller than the right and a number of the right is smaller than the left but with little quantum.

So the left is smaller than the right

So \(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+1}{2010}+\dfrac{x+11}{2000}\)

The average of 4, 15 and 35 is:

\(\dfrac{4+15+35}{3}=\dfrac{54}{3}=18\)

Answer: 18

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

\(=\dfrac{30}{60}+\dfrac{10}{60}+\dfrac{5}{60}+\dfrac{3}{60}+\dfrac{2}{60}\)

\(=\dfrac{30+10+5+3+2}{60}=\dfrac{50}{60}=\dfrac{5}{6}\)

x⁴ = 4x + 1

=> x⁴ - 4x - 1 = 0

=> \(\left(x^2\right)^2+2x^2+1-2x^2-4x-2=0\)

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

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

\(\Rightarrow\left(x^2+1\right)^2=2\left(x+1\right)^2\)

\(\Rightarrow\sqrt{\left(x^2+1\right)^2}=\sqrt{2\left(x+1\right)^2}\)

\(\Rightarrow x^2+1=\sqrt{2}\cdot\left(x+1\right)\)

=> x = 1,663251939

x.y = x:y

=> x.y.y = x

=> xy² = x

=> y² = 1

=> y = \(\pm1\)

* x + y = xy

\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{x}{xy}+\dfrac{y}{xy}=\dfrac{1}{y}+\dfrac{1}{x}=1\)

If y = 1

=> 1/x = 1 - 1 = 0 --> nonsense

If y = -1

=> 1/x = 1 - [-1] = 2

=> 1/x = 2

=> x = 1/2

So \(\left(x,y\right)=\left(\dfrac{1}{2};-1\right)\)

In 1h Ann do the work part number is:

1 : 1 = 1 [job]

In 1h Ben do the work part number is:

1 : 2 = 1/2 [job]

In 1h Ann and Ben do the work part number is:

1 + 1/2 = 3/2 [job]

The two of you do the work that will be done after the time is:

1 : 3/2 = 2/3 [hour]

2/3h = 40minutes

So choose B

1 + 1 = 2

So 1 + 1 written in base 2 is 10 

\(x^2+2x+3\)

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

\(=\left(x+1\right)^2+2\)

We have:

\(\left(x+1\right)^2⋮x+1\)

\(\Rightarrow2⋮x+1\)

\(\Rightarrow x+1\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1\right\}\)

So when x = -2; x = 0; x = -3 or x = 1 

=> \(x^2+2x+3⋮x+1\)

The word "Prove that" is wrong

\(5^x+5^{x+1}=750\)

\(\Rightarrow5^x\left(1+5^1\right)=750\)

\(\Rightarrow5^x\cdot6=750\)

\(\Rightarrow5^x=\dfrac{750}{6}=125\)

\(\Rightarrow x=3\)

\(\dfrac{81^4\cdot3^{10}\cdot27^5:3^{12}}{3^{18}:9^3\cdot243^2}=\dfrac{\left(3^4\right)^4\cdot3^{10}\cdot\left(3^3\right)^5:3^{12}}{3^{18}:\left(3^2\right)^3\cdot\left(3^5\right)^2}\)

\(=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}:3^{12}}{3^{18}:3^6\cdot3^{10}}=\dfrac{3^{16+10+15-12}}{3^{18-6+10}}=\dfrac{3^{29}}{3^{22}}=3^7=2187\)

\(\left(3-4x\right)^3=-125\)

\(\Leftrightarrow\left(3-4x\right)^3=\left(-5\right)^3\)

=> 3 - 4x = -5

=> 4x = 3 - \(\left(-5\right)\)

=> 4x = 8

=> x = 2