A notebook costs: 16.80 : 8 = 2.10 ($)

A pen costs: 15.25 : 5 = 3.05 ($)

Anna buys: 2.10 x 4 + 3.05 x 8 = 8.40 + 24.40 = 32.80 ($) 

Anna received: 2 x 20 - 32.80 = 7.20 ($)

Benny buys: 5 x 2.10 + 10 x 3.05 = 10.50 + 30.50 = 41 ($)

Benny received: 2 x 20 - 41 = -1 ($) 

=> Anna received more and more than: 7.20 + 1 = 8.20 ($)

Because 1 = 10 so 10 = 1 :D 

If x = 2 , we have x² + 2^x = 8 ( scrap because 8 isn't a prime number)

If x = 3 , we have x² + 2^x = 17 ( satisfy )

If x > 3 and x is a prime number then x =3k + 1 or 3k + 2 ( k is a natural number)

+) x = 3k +1

If x = 3k +1 then x² divide to 3 has the remainder 1

and 2^x divide to 3 has the remainder 2

=> 2^x + x² divided to 3 

But 2^x + x² > 3 

So 2^x + x² isn't a prime number

Do the similar to x = 3k+2

The answer is x = 3

This is a problem in the newspaper:"Toán học tuổi thơ" that you must be solve it yourself, shouldn't give it to here!!

60 x 50 = 300 (cm²)

We have 3 straight points: (A;E;B);(B;G;C);(C;D;H);(A;I;D);(A;O;C);(I;O;G);(D;O;B);(E;O;H). So the middle point is: E;G;H;I;O

The number of red marbles is: 15-2-4 = 7(marbles)
The quotion of the red marble is: 7/15

Given: \(a\ne b\ne c\) and \(\left(b-c\right)\sqrt[3]{1-a^3}+\left(c-a\right)\sqrt[3]{1-b^3}+\left(a-b\right)\sqrt[3]{1-c^3}=0\)

Prove: \(\left(1-a^3\right)\left(1-b^3\right)\left(1-c^3\right)=\left(1-abc\right)^3\)

#35# = 35 x 1 x 2 x 3 = 35 x 6 = 210

Call the number of teams is n

We have: The number of games is n(n+1) games

But each game repeated 2 times 

So the number of the games is \(\dfrac{n\left(n+1\right)}{2}\)

There are 21 games

So \(\dfrac{n\left(n+1\right)}{2}=21\Rightarrow n\left(n+1\right)=42\Rightarrow n=6\)

\(\left(x+2y\right)^2+\left(2x-y\right)^2-\left(5x+y\right)\left(x-y\right)-10\left(y+3\right)\left(y-3\right)\)

\(=x^2+4xy+4y^2+4x^2-4xy+y^2-5x^2+5xy-xy+y^2-10\left(y^2-9\right)\)

\(=-4y^2+4xy+90\)

We have:

\(\left(3x-5\right)^{2016}\ge0\) with \(\forall x\in R\)

\(\left(y^2-1\right)^{2008}\ge0\) with \(\forall y\in R\)

\(\left(x-z\right)^{2100}\ge0\) with \(\forall z\in R\)

\(\Rightarrow\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}\ge0\) with \(\forall x,y,z\in R\)

Sign "=" occur 

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\\z=\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left(x;y;z\right)\in\left\{\left(\dfrac{5}{3};1;\dfrac{5}{3}\right);\left(\dfrac{5}{3};-1;\dfrac{5}{3}\right)\right\}\)

Call the number of red marbles is r

the number of blue marbles is b

We have: r + b = 36

and 2r + 6 = b 

=> r + b = r + 2r + 6 = 36

=> r = 10 

We have: 2@6 = ( 6 + 2 )( 6 - 2 ) = 8 . 4 = 32

32@ -4 = ( 32 - 4 )( 32 + 4 ) = 28 . 32 = 896

=> 2. ((2@6)@-4) - 1 = 2. 896 - 1 = 1791

We have: The 4-people pizza's area is: \(8^2\Pi=200,96\left(inch^2\right)\)

So each person can eat: \(\dfrac{200,96}{4}=50,24\left(inch^2\right)\)

=> 16 people eat: \(50,24.16=803,84\left(inch^2\right)\)

=> The suitalbe pizza has the diameter is: \(\sqrt{\dfrac{803,84}{3,14}}.2=32\left(inch\right)\)

This is geometry not algebraic men :) 

Give you a :) 

Call the radius is r, the height is h

We have: 

Cylinder 1 has the volume is: \(\left(2r\right)^2.\Pi.h=\Pi.4r^2.h\)

Cylinder 2 has the volume is: \(r^2.\Pi.k.h\)

We have: \(\dfrac{C1}{C2}=1\)

\(\Rightarrow\dfrac{4r^2.\Pi.h}{r^2.\Pi.k.h}=1\)

\(\Rightarrow\dfrac{4}{k}=1\)

\(\Rightarrow k=4\)

A package of Koka-Kola is: 12 x 8 = 96 (ounces)

A package of Pepsy-Kola is: 16 x 6 = 96 (ounces)

=> The absolute different is 0, not right yet ^^

Call Yumi's age is a

Rana's age is b

Victoria's age is c.

We have: a + b + c = 42 

=> ( a - 4 ) + ( b - 4 ) + c = 36

But a - 4 + b - 4 = c 

=> a + b = c + 8 (1)

We also have: a + b + c = 42

=> a + b = 42 - c (2)

From (1) and (2) we have:

42 - c = c + 8

=> 42 - 8 = 2c

=> 36 = 2c

=> c = 18

So current Yumi's age is 18.

We have: 30 students have lived in Texas less than a year 

=> There are 5% students of 600 students at Goodnight Middle School in Texas have lived here less than 1 year.

So there 65% students are not native Texas lived here less than a year and more than 10 years.

But there are 85% are not native Texas 

=> The non-native students have lived more than 1 year but less than 10 year is 20% = 120(students)

Not plus more 42, because 42 = 40 + 2 

And 2 appear 2 times and 40 appear 2 times => There must be 3(2+3+...+40)

Hey Dao Trong Luan, if you see more strictly, you will see the number 4 appear 3 times, do the same with all the number left, then it must be 3 ( 2+3+...+40) 

We have:

The sum of all the angle in a convex polygon is 180 . ( n - 2 )(degrees) with n = the number of sides

=>180 . ( n - 2 ) = 9720

=> n - 2 = 54

So this is a 52-side polygon.

Is it answered?

Call the number is a, we have:

a = 8k + 7

a = 9k + 8

a = 12k + 11

=> a + 1 \(⋮8;9;12\)

and a is the smallest

So a + 1 = 72

=> a = 71

=> The value of the sum is:

(1+2+3+39+40+41) x 39 / 2 = 2457

We have: 

\(1342_8=1.8^3+3.8^2+4.8+2=738_{10}\)

We have: 

\(2007^{2008}=2007^{4.502}=\left(\overline{...1}\right)^{502}=\overline{...1}\)

\(2008^{2007}=2008^{4.501+3}=\left(\overline{...6}\right)^{501}+2008^3=\overline{...6}+\overline{...2}=\overline{...8}\)

\(\Rightarrow2007^{2008}+2008^{2007}=\overline{...1}+\overline{...8}=\overline{...9}\)

So the ones digit is 9

We have: 2,6,10,14,18,22,26,...

=> x = 18

y = 22

=> x + y = 18 + 22 = 40

We have:

x + 2x + 3x + ... + 99x + 100x = 100

=> 5050x = 100

=> x = \(\dfrac{100}{5050}=\dfrac{2}{101}\)

We have:

\(\dfrac{1}{a}.4=a\)

\(\Rightarrow a^2=4\Rightarrow a\in\left\{2;-2\right\}\)

\(\dfrac{1}{b}.9=b\Rightarrow b\in\left\{3;-3\right\}\)

\(\Rightarrow a+b\in\left\{5;-1;1;-5\right\}\)