Change number 6

We have : 

26 - 63 = 64 - 63 = 1

Answer : 1

(1 + 2 + 3 + ... + 99) + 10

= 4950 + 10 = 4960

3a2 = 2a3

=> a3a2

 = 32 => a = 32

We see that, if x satisfies the equation above => 

−8<3x+4<−32

 and 8<3x+4<32

So we have 28⋅2=56

 possible values of x.

Solve each equations we will have the exact values of x.

450=2100=210.210=(...4).(...4)=...6

So .......

Without loss of generality, we may assume gcd(a,b,c) = 1.

(otherwise, if d=gcd(a,b,c) the for a'=a/d, b'=b/d, c'=c/d, the equation still holds for a', b', c' and a'b'c' is still a cube if only if abc is a cube).

We multiply equation by abc, we have:

  a2c+b2a+c2b=3abc

(*)

if abc=±1

, the problem is solved.

Otherwise, let p be a prime divisor of abc. Since gcd(a,b,c)=1, the (*) implies that p divides exactly two of a, b,c. By symetry, we may assume p divides a, b but not c. Suppose that the lagest powers of p dividing a, b are m, n, respecively.

If n < 2m, then n+1≤2m

  and pn+1| a2c,b2c,3abc. Hence pn+1|c2b, forcing p|c (a contradiction). If n > 2m, then n≥2m+1 and p2m+1|c2b,b2a,3abc. Hence p2m+1|a2c, forcing p|c (a contradicton). Therefore n = 2m and abc=Πp3m, p|abc

, is a cube.

For (x2−1)(x2−4)(x2−7)(x2−10)<0

=> x2−1;x2−4;x2−7;x2−10

 have a positive and 3 negative or a nagative and 3 positive

* If have a positive and 3 negative:

=> x2 - 1 > 0 > x2 - 4 > x2 - 7 > x2 - 10

=> x2 - 1 > 0 and x2 - 4 < 0

=> 1 < x2 < 4 => x2 can equal 2 or 3.

 But x is integer so x2 can't equal 2 or 3

* If have a negative and 3 positive:

=> x2 - 1 > x2 - 4 > x2 - 7 > 0 > x2 - 10

=> x2 - 7 > 0 and x2 - 10 < 0

=> 7 < x2 < 10

=> x2 = 9 because x is integer

=> x = ±3

But x is the positive integer => x = 3

Equivalent disequations:(x4−11x2+10)(x4−11x2+28)<0

Put x4−11x2+10=t

We have:t(t+18)<0

TH1:{t>0t+18<0⇒{t>0t<−18<0

 (loại)

TH2:{t<0t+18>0⇒{t<0t>−18

⇒t∈{−17,−16,...........,−1}

It's easy here, you try it yourself......

For (x2−1)(x2−4)(x2−7)(x2−1)<0

=> In this four numbers, have:

[a positive number and 3 negative numbersa negative number and 3 positive numbers

* If have a positive number and 3 negative numbers:

=> x2-1 > 0 > x2-4 > x2-7 > x2-10

=> x2 - 1 > 0 and x2 < 4

=> 1 < x2 < 4 => No number is satisfy

* If have a negative number and 3 positive numbers:

=> x2-1 > x2-4 > x2-7 > 0 > x2-10

=> x2-7 > 0 and x2-10 < 0

=> 7 < x2 < 10

=> x2 = 9 because x is a positive integer <=> x2 is a positive integer too.

=> x = 3 because x is a positive integer

This is easy. Each number is formed by reading out the digits, grouping digits together and saying the number of identical digits followed by the digit.

1 is read as "one one" (one number 1) -> 11

11 is read as "two ones" (two number 1) -> 21

21 is read as "one two, one one" (one number 2 then one number 1) -> 1211

1211 is read as "one one, one two, two ones" (one number 1, then one number 2, then two number 1 -> 111221.

SO THE NEXT NUMBER IS: "three one, two two, one one" -> 312211

We have:

49 = 7 . 7 ( from the number 77)

36 = 4 . 9 (from the number 49)

18 = 3 . 6 (from the number 36)

SO THE NEXT NUMBER IS: 1 . 8 = 8

We have:

2 = 1 + 1

4 = (1 + 2) + 1

7 = (1 + 2 + 3) + 1

11 = (1 + 2 + 3 + 4) + 1

16 = (1 + 2 + 3 + 4 + 5) + 1

SO THE NEXT NUMBER IS: (1 + 2 + 3 + 4 + 5 + 6) + 1 = 22

P/s: Your question is incorrect

It must be: "FIND THE MISSING NUMBER" or "FIND THE NUBER THAT IS MISSING", not number missing

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)

=12(21.2.3+22.3.4+...+2(n−1)n(n+2))

=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))

=12(12−1n(n+2))

=14.12n(n+2)

⇒123+133+...+1n3<14−12n(n+2)<14

⇒123+133+...+1n3<14

So 123+133+...+1n3<14

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)

=12(21.2.3+22.3.4+...+2(n−1)n(n+2))
=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2)) =12(12−1n(n+2)) =14.12n(n+2) ⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14 So 123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2) =12(21.2.3+22.3.4+...+2(n−1)n(n+2)) =12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2)) =12(12−1n(n+2)) =14.12n(n+2) ⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14 So 123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2) =12(21.2.3+22.3.4+...+2(n−1)n(n+2)) =12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2)) =12(12−1n(n+2)) =14.12n(n+2) ⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14 So 123+133+...+1n3<14

63×84=2×2=4

38×56=1548=516

10812×881=9×881=89

123×133688=16,359688

410=40100=40%

23100=23%

2541000=25.4100=25.4%

789010000=78.9100=78.9%

The area of the square is: 38⋅38=382=1444(m2)

If you rearrange it, the sum of the digits is still 15, which makes it impossible to rearrange the digits of 3255 to make a number that is not a multiple of 3

But i also have the answer

           The area of the rectangle is :

                    8 x 9 = 72 ( cm2 )

                              Answer : 72 cm2

We only need two number: 17 and 16

(17 + 16 + 17) + (17 + 16 + 17)

AUTO ANSWER (AFTER SEVERAL DAYS): 

This is not a hard question, but many can't solve it. Because they thought: "The question is wrong!". Exactly, 99 = 72 + 27; 45 = 27 + 18; 39 = 18 + 21; So the ?? must be 15. But why does 21 = 13 + 7? We have gone into the wrong MINDWAY.  Let's think of another way. We have 9 + 9 + 7 + 2 = 27 (Sum of the digits in line 1, row 1 + sum of the digits in line 2, row 1 = line 2, row 2). Because 2 + 8 + 1 + 2 = 13 and 3 + 6 + 2 + 1 = 12 so the answer correct is 12. 

(#560 - 1000 best IQ questions in history)

We have: 1 + 5 = 2 + 4 = 6 (there are no 2 number-3 cards)

=> A has (1;5) or (2;4) cards [1]

B said the difference between two cards is 5

=> B has (1;6), (2;7), (3;8) or (4;9) cards [2]

We have: 2*9 = 3*6 = 18

=> C has (2;9) or (3;6) cards [3]

We have: 2 = 2*1; 4 = 2*2; 6 = 2*3; 8 = 2*4

=> D has (1;2), (2;4), (3;6) or (4;8) cards [4]

If A has (2;4) cards => Unsatisfy

If A has (2;4) cards -> B has (1;6) or (3;8) cards

 - If B has (1;6) cards => Unsatisfy for C

 - If B has (3;8) cards => Unsatisfy for C, too

=> A has (1;5) cards

=> B has (2;7), (3;8) or (4;9) cards

- If B has (2;7) cards -> C has (3;6) cards -> D has (4;8) cards -> The number of the remaining card is 9.

- If B has (3;8) cards -> C has (2;9) cards -> Unsatisfy for D

- If B has (4;9) cards -> C has (3;6) cards -> Unsatisfy for D

So the number of the remaining card is 9

m.n=(999...99).(888...88)

                2015            2015

         =(888...88).(1000...00−1)

                2015             2016

        =(888...88000..00)−(888...88)

                     4030                      2015

        =888...887111...112

               2014         2014

Calculate \(p=\dfrac{3x-2y}{3x+2y}\) know x , y satisfy x < 2y and  \(9x^2+4y^2=20xy\)

Let x , y , z > 0 .Validity : \(p=\dfrac{x^{2018}+1}{x^{2018}+y^{2018}+z^{2018}+3}\)  know  \(x^3+y^3+z^3=3xyz\)

Let a, b, c be other numbers 0 satisfying : \(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\\a^3+b^3+c^3=2^9\end{matrix}\right.\)

Calculate  \(p=a^{2019}+b^{2019}+c^{2019}\)

Let a , b , c be positive numbers . Prove that : \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{1}{2}\left(a+b+c\right)\)

Let x , y , z be positive real numbers . Prove that \(\dfrac{x^2-z^2}{y+z}+\dfrac{y^2-x^2}{z+x}+\dfrac{z^2-y^2}{x+y}\ge0\)

Let a , b, c be a positive real numbers , prove that :

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{1}{\sqrt{2}}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\)

What is the condition for the equality ?

thank you

Solve the following system of equations :

\(\left\{{}\begin{matrix}4\sqrt{3x+4y}+\sqrt{8-x+y}=23\\3\sqrt{8-x+y}-2\sqrt{38+6x-13y}=5\end{matrix}\right.\)

thank you

Given the square triangle ABC at A , the AD curve ( D of BC ) .

Prove that \(\dfrac{\sqrt{2}}{AD}=\dfrac{1}{AB}+\dfrac{1}{AC}\)

A trapezoid has an area of 1 . Ask the diagonal of this trapezoid is the smallest .

For the acute triangle ABC , the circle (O) has a fixed BC ( BC ≠ 2R ) , H is the center. Determine the position of A to: S = HA + HB + HC is the maximum value.

Given the following equation :

\(2x^4-4x^3+\left(4-a\right)x^2+\left(a-2\right)x+a-a^2=0\)

Prove that the equation has only one negative root when \(a>1\)

A NEW YEAR AFTER KHANG THINH VUONG 2018

YOU'RE IN TALKING TO XUAN SANG
THE MOMENT OF SMALL COMMUNITY

WE HAVE COME BACK TO 2018

ONE YEAR ANNIVERSARY OF 2018 PROFESSIONAL WAY

YEAR OF THE YEAR 2018

HAPPY NEW YEAR .

WELCOME TO YOU THAT YOU LIKE HAVING A GREAT YEAR, GOOD LESS THANKS. "HELP YOU SOLVE MATH" THAT WAS WRONG.

Solve the following equation with the parameter m :

\(\dfrac{x+1}{x-m}+\dfrac{x+m}{x-1}=2\)

solve the equation :

\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)

 

solving the equation with a , b , c is the parameter : ( abc is different 0 )

\(\dfrac{x-b-c}{a}+\dfrac{x-a-c}{b}+\dfrac{x-a-b}{c}=3\)