We have :

\(\dfrac{1}{1^2}>\dfrac{1}{1.2}\)

\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

..............

\(\dfrac{1}{100^2}>\dfrac{1}{99.100}\)

=> \(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}=A\)

So B > A

x⁴ + x³ + 2x² + x + 1 = 0

x³.(x + 1) + x(x + 1) + x² + 1 = 0

(x³ + x)(x + 1) + x² + 1=  0

x(x² + 1)(x + 1) + x² + 1 = 0

Give x² + 1 = t

=> x(x + 1) . t + t = 0

=> [x(x + 1) + 1] . t = 0

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)+1=0\\t=0\end{matrix}\right.\)

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

Right here, i can't do anything more , is there anyone can do more ???

x = 3y

=> \(\dfrac{1}{x}=\dfrac{1}{3y}\)

We got this :

\(\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{\dfrac{1}{x}-\dfrac{1}{y}}=\dfrac{\dfrac{1}{3y}+\dfrac{1}{y}}{\dfrac{1}{3y}-\dfrac{1}{y}}=\dfrac{\dfrac{4y}{3y^2}}{\dfrac{-2y}{3y^2}}=\dfrac{4y}{3y^2}.\dfrac{3y^2}{-2y}=\dfrac{4y}{-2y}=-2\)

You have to prove this :

x³ + y³ + z³ \(\ge\) 3xyz

1) We have :

x³ + y³ + z³ \(\ge\) 3xyz

<=> x³ + y³ + z³ - 3xyz \(\ge\) 0

Now , we are going to prove this :

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

You can prove that :) 

So , we have :

 (x + y + z)(x² + y² + z² - xy - yz - zx) \(\ge\) 0

Suppose, this inequality is true .

In this inequality , we have two case .

1. Two factor are positive 

2. Two factor are negative 

In the second case , it can not take place 

Because x,y,z are positive mean x + y + z \(\ge\) 0 (1)

So , the second case is failed .

=> x² + y² + z² - xy - yz - zx \(\ge\) 0

=> 2x² + 2y² + 2z² - 2xy - 2yz - 2zx \(\ge\) 0

=> (x - y)² + (y - z)² + (z - x)² \(\ge\) 0 (right)      (2)

From (1) and (2)

=> We have things to prove .

\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}+\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

Because BD = DC => D is the midpoint of BC

Other way , AD cut BC at midpoint D

=> AD is median line of \(\Delta ABC\)

With AD = DC = AD 

=> \(\Delta ABC\) is square triangle (Because of the nature of the median lines in the triangle)

Apply properties of equal ratio ranges , we have :

\(\dfrac{a+b-c}{c}=\dfrac{a+b-c}{b}=\dfrac{c+b-a}{a}=\dfrac{a+b-c+a+b-c+c+b-a}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=2c\\a+c-b=2b\\b+c-a=2a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=3c\\a+c=3b\\b+c=3a\end{matrix}\right.\)

Change all these equalities into x , we have :

x = \(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{3c.3a.3b}{abc}=3\)

Hence , x = 3 

Call three fractions is A,B,C

We have : \(A=\dfrac{x}{y}\) ; \(B=\dfrac{y}{t}\) ; \(C=\dfrac{e}{f}\)

Follows the threads : \(\dfrac{x}{3}=\dfrac{z}{4}=\dfrac{e}{5}\) ; \(\dfrac{y}{5}=\dfrac{t}{1}=\dfrac{f}{2}\)

\(\Rightarrow\dfrac{\dfrac{x}{3}}{\dfrac{y}{5}}=\dfrac{\dfrac{z}{4}}{\dfrac{t}{1}}=\dfrac{\dfrac{e}{5}}{\dfrac{f}{2}}\)

\(\Rightarrow\dfrac{x}{3}.\dfrac{5}{y}=\dfrac{z}{4}.\dfrac{1}{t}=\dfrac{e}{5}.\dfrac{2}{f}\)

\(\Rightarrow\dfrac{x}{y}.\dfrac{5}{3}=\dfrac{z}{t}.\dfrac{1}{4}=\dfrac{e}{f}.\dfrac{2}{5}\)

\(\Rightarrow A:\dfrac{3}{5}=B:4=C:\dfrac{5}{2}\)

Applying the same sequence properties we have:

\(\dfrac{A}{\dfrac{3}{5}}=\dfrac{B}{4}=\dfrac{C}{\dfrac{5}{2}}=\dfrac{A+B+C}{\dfrac{3}{5}+4+\dfrac{5}{2}}=\dfrac{\dfrac{213}{70}}{\dfrac{71}{10}}=\dfrac{3}{7}\)

=> \(\left\{{}\begin{matrix}A=\dfrac{3}{7}.\dfrac{3}{5}=\dfrac{9}{35}\\B=\dfrac{4.3}{7}=\dfrac{12}{7}\\C=\dfrac{5}{2}.\dfrac{3}{7}=\dfrac{15}{14}\end{matrix}\right.\)

The answer is :

(12 x 3) + (4 x 2) = 36 + 8 = 44

In 1 tetrahedrons , we have 4 faces around it

With 1 cube , we have 6 faces 

That so , if we have 13 tetrahedrons mean we have 4.13 = 52 faces

And with 5 cubes mean we have 5.6 = 30 faces

The total is : 52 + 30 = 82 faces 

a) Apply Inequality - Cauchy , we have :

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2.\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2.1=2\)

b) We have :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\) (Đúng)

=> \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

1. Call two consecutive odd number is :

2k + 1 ; 2k + 3

We have :

\(\dfrac{\left(2k+1\right)+\left(2k+3\right)}{2}=248\)

=> 4k + 4 = 2.248

=> 2k + 2 = 248

=> 2k =  246

=> k = 123

So , two consecutive odd number is :

2k + 1 = 2.123 + 1 = 246 + 1 = 247

2k + 3 = 2,123 + 3 = 246 + 3 = 249 

Follows the picture , we can see area A , B and C are square .

Call the length of the square edges of squares A, B, C in turn a, b, c

=> a = \(\sqrt{S_A}=\sqrt{100}=10\) ; b = \(\sqrt{S_B}=\sqrt{36}=6\)

Apply the Pythagoras , we have :

b² + c² = a²

c² = a² - b²

c² = 10² - 6² = 64

=> c = 8 (c can not be equal to -8 because it's the length of an edge)

=> S_C = c² = 8² = 64 (cm²)

We have : 3.5.7.9.11 = (3.5).7.9.11 = 15.7.9.11 = .......5

So , the last digit number is 5

We have :

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{2017^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2016.2017}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2016}-\dfrac{1}{2017}=1-\dfrac{1}{2017}< 1\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{2017^2}< 1\)

We have :

\(\widehat{A}=3\widehat{B}\) => \(\widehat{B}=\dfrac{\widehat{A}}{3}\)  (1)

\(2\widehat{A}=\widehat{C}\)  (2)

Apply a total of 3 corners in a triangle :

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

Change (1) and (2) in this equality :

=> \(\widehat{A}+\dfrac{\widehat{A}}{3}+2\widehat{A}=180^0\)

=> \(\dfrac{10}{3}\widehat{A}=180^0\)

=> \(\widehat{A}=54^0\)

\(\left(5\dfrac{1}{3}-2\dfrac{1}{2}\right)+\left(5\dfrac{1}{2}-3\dfrac{1}{3}\right)\)

\(=5\dfrac{1}{3}-2\dfrac{1}{2}+5\dfrac{1}{2}-3\dfrac{1}{3}\)

\(=\left(5\dfrac{1}{3}-3\dfrac{1}{3}\right)+\left(5\dfrac{1}{2}-2\dfrac{1}{2}\right)\)

\(=2+3=5\)

Phan Thanh Tinh  ok ok , i'm sorry !

13 = 1.13

4 = 2² 

6 = 2.3

BCNN(13 ; 4 ; 6) = 13.2².3 = 156