Prove that \[ \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{20}} \] is a natural number.

Find the function \(f(x)\) satisfying: \[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Find the function \(f(x)\) satisfying:

\[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Solution:

Put \(y=1\), and let \(f(1)=a\).

\[ af(x)-f(x)-90=\frac{10(x+1)}{x} \]

\[ (a-1)f(x)=100+\frac{10}{x} \]

Therefore,

\[ f(x)=A+\frac{B}{x} \]

 

Math PreyVeng Cambodia 2023

• Day 01

Math Chum Kiri District, Kampot 24/03/2023

• Solution Here:  Download PDF

Math Out Standing 2023 Provincial, Cambodia

Solution 

If \(x_1,x_2\) are the roots of the equation \[ x^2-x-3=0 \] Find the value of \[ A=7x_1^5+19x_2^4 \]

Solution:

Since \(x_1,x_2\) are roots of

\[ x^2-x-3=0 \]

we have:

\[ x^2=x+3 \]

For any root \(x\):

\[ x^3=x(x^2)=x(x+3)=x^2+3x=4x+3 \]

\[ x^4=x(4x+3)=4x^2+3x=7x+12 \]

\[ x^5=x(7x+12)=7x^2+12x=19x+21 \]

Math Problem and Solution Video

 

Problem 000

Prove that \[ a^2+b^2+c^2\geq a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}} \]

Prove that \[ x^{2023}+y^{2023}+z^{2023}=0 \] If \[ \frac{x^2+y^2+z^2}{a^2+b^2+c^2} = \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \]

• This is the problem that I picked from Vietnamese collection problems which is the basic for kind of this question. 
• From that origin problem is `x^2019+y^2019+z^2019=0` but, I have changed it to `2023` .

• Problem in Khmer language

• Proof:

  Let

  \[ A=a^2+b^2+c^2 \]

  By the Cauchy-Schwarz inequality:

  \[ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) (a^2+b^2+c^2) \geq (x+y+z)^2 \]

  Given that

  \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = \frac{x^2+y^2+z^2}{a^2+b^2+c^2} \]

  we obtain

  \[ x^2+y^2+z^2\geq(x+y+z)^2 \]

  Expanding:

  \[ x^2+y^2+z^2 \geq x^2+y^2+z^2+2xy+2yz+2zx \]

  Therefore:

  \[ xy+yz+zx\leq0 \]

  Since the condition forces the symmetric relation, we get:

  \[ x+y+z=0 \]

  Hence:

  \[ (x+y+z)(x^{2022}-x^{2021}y+\cdots+y^{2022}) \]

  gives

  \[ x^{2023}+y^{2023}+z^{2023}=0 \]

  Therefore:

  \[ \boxed{x^{2023}+y^{2023}+z^{2023}=0} \]

  Click here to download PDF file: Download Here

• Another problem, you all can learn more.

Prove that if \(x,y,z>0\), then \[ M=\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z} \leq\frac34 \]

 

Find all number which its square has 4 digits number and divisible 33

• In order to solve type of this problem, you have to understand about divisible theory of natural number. As GCD (Greatest Common Divisor) and LCM (Least Common Divisor). When you know about GCD and LCM, you surely can simplify the problem to be more easier to solve. 
• As example of our problem, `33=3.11` and `GCD(3;11)=1` Then, our problem can be found as divisible with `3` and `11` 

 Solution

Solve the equation: \[ \sin^{2012}x+\cos^{2012}x=\frac{1}{2^{1005}} \]

Solve the equation `sin^2012x+cos^2012x=1/2^1005`

• I will solve this problem  by Khmer language and in order to solve this problem, you have to use derivative of function which you call find the minimum value of function .
• This is the problem 02 of Vietnamese Out-Standing Student grade 12.

Vietnamese Mathematics Provincial 2011

Vietnamese Mathematics Provincial 2011

• This is the problems of Vietnamese Mathematics Provincial examination in 2011. I just picked problem number 04 to share all of you which is related to sequence. As you know, sequences are the problems which need more strategy to solve where you have to combine all your understanding.

Cambodia National Math 2019, 22/04/2019 Day 02

 

Cambodia National Math 2019, 22/04/2019 Day 02

Math Cambodia 2019 Day 02

• This was the problem that released for Out Standing Student in Cambodia in 2019 for 2nd day of testing.
• There were two days of the testing. This is the first day of exam.
• You all can Click here to download

Cambodia Grade 12, 22/04/2019 Day 01

Cambodia National Math 2019, 22/04/2019

• This was the problem that released for Out Standing Student in Cambodia in 2019.
• There were two days of the testing. This is the first day of exam.

• You can watch solution here:

Mathematics Out Standing Student Phnom Penh 2020, Cambodia

 Mathematics Out Standing Student Phnom Penh 2020, Cambodia

• This was the problem that released for Out Standing Student in Phnom Penh, Cambodia in 2020.
• There were two days of the testing. This is the first day of exam.

Prove that the polynomial \[ x^{9999}+x^{8888}+x^{7777}+\cdots+x^{1111}+1 \] is divisible by \[ x^9+x^8+x^7+\cdots+x+1 \]

 Problem: 01

Prove that the polynomial `x^9999+x^8888+x^7777+...+x^1111+1` is divisible by `x^9+x^8+x^7+....+x+1`

Solution

Find all functions \(f(x)\) satisfying: \[ (x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2) \]

 

Find all functions \(f(x)\) satisfying:

\[ (x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2) \]

Solution

Problem If $$ x+\frac{1}{x}=2, $$ find the value of $$ x^5+\frac{1}{x^5}. $$

 If \x+1/x=2\ Find the value of \x^5+1/x^5\

As we had: `x+1/x=2` `rightarrow(x+1/x)^2=4` `leftrightarrowx^2+1/x^2=2`

We continue with `(x^2+1/x^2)(x+1/x)=4` `leftrightarrowx^3+x+1/x+1/x^3=4`

                                `leftrightarrowx^3+1/x^3=2`

Vietnamese Mathematical Olympiad 2022 Let \(a\), \(b\), and \(c\) be the roots of the equation $$ x^3-x^2-1=0. $$ Find the value of $$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Let \(a\), \(b\), and \(c\) be nonzero real numbers satisfying

$$ a+b+c=2022, $$ and $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}. $$

Find the value of

$$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Solution