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hghfghfgh 26/03/2017 at 20:16

At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

+ if a odd and b even, or a even and b odd then a + b or a - b is still odd

+ if a and b are both even then a + b or a -b is still even

+ If a and are both odd then a + b or a - b is even

So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).
Faded 19/01/2018 at 14:52

At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

+ if a odd and b even, or a even and b odd then a + b or a - b is still odd

+ if a and b are both even then a + b or a -b is still even

+ If a and are both odd then a + b or a - b is even

So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even). [haha]
An Duong 10/03/2017 at 14:14

At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

+ if a odd and b even, or a even and b odd then a + b or a - b is still odd

+ if a and b are both even then a + b or a -b is still even

+ If a and are both odd then a + b or a - b is even

So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).

triệu lệ dĩnh 08/07/2017 at 10:07

30 x 40 = 1200

tcik nha
Nguyễn Thị Ý Nhi selected this answer.
Trần Quỳnh Anh 27/12/2017 at 19:15

30.40

=3.10.4.10

=(3.4).(10.10)

=12.100

=1200

So the answer is 1200
NGUYỄN SANH KIÊN 08/07/2017 at 11:30

1200

good luck !

jimin bts cute 27/07/2018 at 02:01

Z ={.....-3,-2.-1,0,1,2,3...}

vy nguyen mini selected this answer.

Nguyễn Đức Kiên 06/04/2017 at 17:15

đúng là đề sai thật rồi . Nhìn nó cứ kì kì làm sao đấy .
Vũ Hà Vy Anh selected this answer.
Phan Thanh Tinh Coordinator 06/05/2017 at 09:53

Draw the altitudes AE,MF,NG as shown

$\dfrac{S_{\Delta ABN}}{S_{\Delta BMN}}=\dfrac{AB.NG}{2}:\dfrac{BM.NC}{2}=BC^2:\dfrac{BC}{2}:\dfrac{CD}{2}=4$

$\Delta ABN,\Delta BMN$ also have the common base BN,so AE = 4MF

$\Delta ABO,\Delta BMO$ have the common base BO and the altitudes AE = 4MF,so $S_{\Delta ABO}=4S_{\Delta BMO}$

$\Rightarrow S_{\Delta ABO}=\dfrac{4}{1+4}\left(S_{\Delta ABO}+S_{\Delta BMO}\right)=\dfrac{4}{5}.S_{\Delta ABM}$

$=\dfrac{4}{5}.\dfrac{20.\left(20:2\right)}{2}=80$(cm²)

$\Rightarrow S_{AOND}=S_{ABCD}-S_{\Delta ABO}-S_{\Delta BNC}$

= 20.20 - 80 - $\dfrac{20.\left(20:2\right)}{2}$ = 220 (cm²)
Vũ Hà Vy Anh 06/04/2017 at 17:17

ai bảo mình mới thi cấp thành phố hôm thứ 6 tuần trước nữa xong đó còn được cầm đề về mà

tth 30/10/2017 at 15:14

We have:

25 do not divide to 10

So 25 can't be true about number N of doors .

Note: I'm not certain
Ma Kết 05/04/2018 at 14:42

We have:

25 do not divide to 10

So 25 can't be true about number N of doors .

Note: I'm not certain
Ma Kết 05/04/2018 at 14:41

Hà Đức Thọ

Phan Minh Anh 10/06/2017 at 15:57

75+25+13+91+87+52+48+9

=(75+25)+(13+87)+(91+9)+(52+48)

= 100 + 100 + 100 + 100

= 100x4

= 400.
Help you solve math 14/08/2017 at 08:45

75+25+13+91+87+52+48+9

=(52+25)+(13+87)+(91+9)+(52+48)

=1000+100+100+100=100x4

=400

mk
Vũ Hà Vy Anh 12/06/2017 at 07:50

75 + 25 + 13 + 91 + 87 + 52 + 48 + 9

= ( 75 + 25 ) + ( 13 + 87 ) + ( 91 + 9 ) + ( 52 + 48 )

= 100 + 100 + 100 + 100

= 100 . 4

= 400

Tokisaki Kurumi 26/04/2017 at 11:47

The finite numbers are 0,2,4,6,8
Nguyễn Thị Thanh Hiền 08/05/2018 at 06:45

Numbers divisible by 2 have the last digits of 0, 2, 4, 6, 8
Nguyen Huu Ai Linh 10/12/2017 at 07:16

The even number or The last right is all these number:" 0; 2; 4; 6; 8."

mathlove 12/03/2017 at 12:37

Suppose that the triangle ABC has A = 90⁰, B = 60⁰, AB = 4. We construct the rectangle ABCD with center O. By the assumption B = 60⁰ , assune that OAB is a equilateral triangle; OB = AB = 4, BC = 8. By the theorem Pytagor we have

$AC=\sqrt{BC^2-AB^2}=\sqrt{8^2-4^2}=4\sqrt{3}$. So $BC=8,CA=4\sqrt{3}$.

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Kudo Shinichi 22/03/2017 at 22:13

It so good for me in the future , thanks you :))

Phan Minh Anh 10/06/2017 at 16:16

There are four ducks.

Ai tk cho mk thì mk sẽ tk lại!
Trần Đức Huy 15/05/2018 at 08:21

there are four ducks
Nguyễn Thị Thanh Hiền 08/05/2018 at 06:47

there are four ducks

Phan Minh Anh 14/06/2017 at 12:58

It Thai Son mountain.
Đỗ Thanh Hải 29/06/2017 at 09:51

Núi Thái Sơn nha Bạn
Phan Minh Anh 10/06/2017 at 16:21

Sorry: What the mountains cut into pieces?

FA KAKALOTS 28/01/2018 at 22:08

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

⇒1n3<1(n−1)n(n+1)

1(n−1)n(n+1)=1n.1(n−1)(n+1)

=1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

=12.(1(n−1)n−1n(n+1))

Now we have :

E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

=12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

=12(12.3−1n(n+1))=112−12n(n+1)<112

Hence,E<112
Phan Thanh Tinh Coordinator 24/04/2017 at 13:50

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n² - 1) = n³ - n < n³

$\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}$

$\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}$

$=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)$

$=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)$

Now we have :

E < $\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}$

$=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}$

Hence,$E< \dfrac{1}{12}$

FA Liên Quân Garena 08/01/2018 at 21:09

We have :

$=\dfrac{\left(x^3+x^2\right)-\left(x+1\right)}{x^2-5x-x+5}$

$=\dfrac{x^2\left(x+1\right)-\left(x+1\right)}{x\left(x-1\right)-5\left(x-1\right)}$

$=\dfrac{\left(x-1\right)\left(x+1\right)^2}{\left(x-5\right)\left(x-1\right)}$

$=\dfrac{\left(x+1\right)^2}{x-5}$

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Thao Dola 14/03/2017 at 14:23

The first such triple is 8 = $2^2+2^2$,9 = $3^3+0^2$,10=$3^2+1^2$, which suggests we consider triples $x^2-1,x^2,x^2+1$.Since $x^2-2y^2=1$ has infinitely many positive solutions (x,y), we see that $x^2-1=y^2+y^2,x^2=x^2+0^2$and $x^2+1$ satisfy the requiment and there are infinitely many such triples.
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FA KAKALOTS 28/01/2018 at 22:12

The first such triple is 8 = 22+22,9 = 33+02,10=32+12, which suggests we consider triples x2−1,x2,x2+1.Since x2−2y2=1 has infinitely many positive solutions (x,y), we see that x2−1=y2+y2,x2=x2+02and x2+1 satisfy the requiment and there are infinitely many such triples.
Such doge 14/03/2017 at 21:03

Wowe it hard

Dao Trong Luan Coordinator 31/12/2017 at 10:33

a.

$x^2-2y^2=xy$

$\Leftrightarrow x^2-2y^2-xy=0$

$\Leftrightarrow\left(x^2-y^2\right)-\left(y^2+xy\right)=0$

$\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0$

$\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0$

$\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0$

But $x+y\ne0$

$\Rightarrow x-2y=0\Leftrightarrow x=2y$

$\Rightarrow P=\dfrac{x-y}{x+y}=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}$

Alone 31/12/2017 at 11:19

Continue Dao Trong Luan'answer:

$\left(x-1\right)\left[x^2-4\left(x-1\right)\right]$

$=\left(x-1\right)\left(x^2-4x+4\right)$

$=\left(x-1\right)\left(x-2\right)^2$
FA Liên Quân Garena selected this answer.
FA Liên Quân Garena 01/01/2018 at 10:26

I edited the subject

$x^{3} - x^{2} - 4 x^{2} + 8 x - 4$

$= x^{2} (x - 1) - (4 x^{2} - 8 x + 4)$

$= x^{2} (x - 1) - [{(2 x)}^{2} - 2 \cdot 2 x \cdot 2 + 2^{2}]$

$= x^{2} (x - 1) - {(2 x - 2)}^{2}$

$= x^{2} (x - 1) - 4 {(x - 1)}^{2}$

$= (x - 1) [x^{2} - 4 (x - 1)]$

$(x - 1) [x^{2} - 4 (x - 1)]$

$= (x - 1) (x^{2} - 4 x + 4)$

$= (x - 1) {(x - 2)}^{2}$
Hương Yêu Dấu 31/12/2017 at 13:33

We have :

(x - 1) . [x² - 4 . (x - 1)]

<=> (x - 1) . (x² - 4x + 4)

=> (x - 1). (x - 2)²

This is brief

Alone 31/12/2017 at 11:07

We have:$\left(x+1\right)\left(x-3\right)-\left(x+5\right)\left(x-5\right)\left(x-2\right)=0$

$\Leftrightarrow x^2-2x-3-\left(x^2-25\right)\left(x-2\right)=0$

$\Leftrightarrow x^2-2x-3-x^3+2x^2+25x-50=0$

$\Leftrightarrow3x^2-x^3+23x-53=0$

$\Leftrightarrow x^2\left(3-x\right)-23\left(3-x\right)+16=0$

$\Leftrightarrow\left(x^2-23\right)\left(3-x\right)+16=0$

$\Rightarrow x^2-23\in\left\{-16,-8,-4,-2,-1,1,2,4,8,16\right\}$

$\Rightarrow x^2\in\left\{7,15,19,21,22,24,25,27,31,39\right\}$

Because 3-x is a integer number so x is a integer number so $x^2=25$ and 3-x=-8

$\Rightarrow$ x=$\pm$5 and x=11 (unsatisfactory)

So not have x satisfy
FA Liên Quân Garena selected this answer.

Lightning Farron 30/12/2017 at 23:44

Stop Touching YourSelf, please =((

Ngu Ngu Ngu 13/04/2017 at 22:55

Put the equation above is $\left(1\right)$

If $xy>0$ then:

$\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{y}+\dfrac{2}{x}=2011\\\dfrac{1}{y}-\dfrac{2}{x}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1007}{9}\\\dfrac{1}{x}=\dfrac{490}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{490}\\y=\dfrac{9}{1007}\end{matrix}\right.$ (satisfy)

If $xy< 0$ then:

$\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{y}+\dfrac{2}{x}=-2011\\\dfrac{1}{y}-\dfrac{2}{x}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-\dfrac{1004}{9}\\\dfrac{1}{x}=-\dfrac{1031}{18}\end{matrix}\right.$$\Rightarrow xy>0$ (unsatisfactory)

If $xy=0$ then: $\left(1\right)\Leftrightarrow x=y=0$ (satisfy)

Conclude: equations have 2 solutions: $\left(0;0\right)$ and $\left(\dfrac{9}{490};\dfrac{9}{1007}\right)$
Nguyễn Thị Huyền Mai selected this answer.

Ngu Ngu Ngu 14/04/2017 at 08:08

$2\left(x^2+\dfrac{1}{x^2}\right)+3\left(x+\dfrac{1}{x}\right)-16=0\left(1\right)$

Condition: $x\ne0$

Put $t=x+\dfrac{1}{x}\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$

$\left(1\right)\Leftrightarrow2t^2+3t-20=0$ $\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=\dfrac{5}{2}\end{matrix}\right.$

If $t=-4\Rightarrow x=-2\pm\sqrt{3}$

If $t=\dfrac{5}{2}$ $\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.$

Conclude:...
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