In the rectangle ABCD, segement AB is twice segement BC. The perimeter of the rectangle  ABCD  is 60cm. Find the area of the rectangle ABCD?

simple expresson:

a,  \(x+22+\left(-14\right)+52\)

b,  \(\left(-90\right)+\left(p+10\right)+100\)

Find x , y , z

a, \(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}=0\)

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 S (x ; y) = x² - y² with real numbers x and y then S (3;S ( 3 ; 4 )) = ...

Suppose a regular hexagon has a perimeter equal to the circumference of a circle. What is the ratio of a side of the hexagon to the radius of the circle? Express your answer as a common fraction in terms of π

Show geometrically (like in my solution) that for positive reals ,

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

16) Given polynomial f(x) that satisfy: f(x+1) = f(x) + 1 (\(x\in R\)) and f(0) = 1

Find f(x)

Find X :

X * 45 + X * 34 + X * 45 

Find number of integers  that satisfy .

In right pyramid PABCD, ABCD is a square with sides of length 2 and the distance from P to the center of square ABCD is 1. A plane intersects the pyramid at A and intersects edges PB, PC and PD at points B', C' and D', respectively. If and , what is the value of the ratio ? Express your answer as a common fraction.

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 .......

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

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)

The minimum value of the expression A= 

Calculate: \(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}=...\)

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

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