There are 2 cases
Cases 1 : The number 3 digit number does not contain the number 3,4,5 in it
Cases 2 :  contains one or two digits

 Case 1 or 2

Because a, b is positive and a² - b² = 144 => 0 < a , b < 12

Instead a = 1 ; 2 ; 3 ; ... ; 11

We see no value of a that satisfies b positive integers

=> No pair satisfy

We have:

K = 2⁵ . 3³ . 5² . 7

The positive integers of K are : 

(5 + 1) . (3 + 1) . (2 + 1) . (1 + 1) = 144 to wish

Answer : ...

We have:

 x – 2xy + y – 3 = 0

<=> 2x – 4xy + 2y – 6 = 0 <=> 2x – 4xy + 2y – 1 =5

<=> 2x(1 – 2y) – (1 – 2y) = 5 <=> (2x – 1)(1 – 2y) =5

Tabulator : You self make 

=> So \(x\in\left\{\left(1;-2\right),\left(3;0\right),\left(-2;1\right)\right\}\)

C1 : Lê Quốc Trần Anh and Đào Trọng Luân

C2:

There are : 100% - 25% = 75% (students) are good students

There are all good students : 900 . 75% = 675 (students) are good students

We have : x² = 6400

=> x² = \(\pm80^2\)

To try on:

- x = 80 => x it isn't a square number

- x = -80 < 0 it isn't square number

=> x it isn't square number

B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

We have:

4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4

= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n - 1)n(n + 1)(n + 2) - [(n - 2)(n - 1)n(n + 1)]

= (n - 1)n(n + 1)(n + 2) - 0.1.2.3 = (n - 1)n(n + 1)(n + 2)

=>B = \(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

 My younger brother marked  I have advised, but now I far away so I can not speak directly with my brother
I am sorry

It should not be marked that anymore

a + b + c = 0

=> (a + b + c)²   \(>_-\)0

<=> a² + b² + c² + 2ab + 2cb + 2ac = 0

=> 2 (ab + bc + ca)   \(< _-\) 0

We have: 

\(\left|\dfrac{2}{7}-\dfrac{1}{6}+x\right|>_-0\)

and \(-\dfrac{9}{4}-\left|y\right|< _-0\)

=> x, y does not exist

Sorry: 

We have : \(\left|x-\dfrac{7}{5}\right|+\left|y+\dfrac{9}{25}\right|=0\)

=> \(\left|x-\dfrac{7}{5}\right|=0;\left|y+\dfrac{9}{25}\right|=0\)

=> \(x=\dfrac{7}{5};y=\dfrac{-9}{25}\)

We have:

\(\left|\dfrac{2}{7}-\dfrac{1}{6}+x\right|>_-0\)

And : \(-\dfrac{9}{4}-\left|y\right|< 0\)

=> x, y does not exist

34² + 66² + 68 . 66

= (34² + 66² + 2 . 34 . 66)

= (34 + 36)² 

=100²

=> 10000

B = (4 - x)² + 5

We have: (4 - x)² > or = 0

=> (4 - x)² + 5 > or = 5

<=> 4 - x = 0

=> x = 4

So B reaches GTNN = 5 when x = 4

Question how new is true : (4 - x)² + 5

When dividing 5 by 3 what is left over?

B, 2 

Because, 5 split 3 residual 2

B, regularly

You should not post such a question on this page

This is not a math problem

P/s : This is an website Mathematics
Not about literature
 

We have :

X * 45 + X * 34 + X * 45

= X * ( 45 + 34 + 45 )

= X * 134

So X can not be worth being 

P/s : Invalid, wrong

Cho mình nói tiếng việt:

Mình nghĩ là các dấu chấm (...) là số cần điền vào xin lỗi mọi người

P/s : wrong topic