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Asuna Yuuki
09/03/2017 at 21:33
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3
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Express 3000 cm3 in liter.

 

  • ...
    Kurosaki Akatsu Coordinator 09/03/2017 at 21:35

    3000cm3 = 3 litter

    Asuna Yuuki selected this answer.
  • ...
    Kurosaki Akatsu Coordinator 09/03/2017 at 21:35

    3000cm3 = 30 litter

  • ...
    FA FIFA Club World Cup 2018 16/01/2018 at 22:00

    3000cm3 = 3 litter


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Bùi Vân
27/04/2017 at 11:32
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0
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Professor Derring's desk has 4 history books, 5 classical fictions and 3 lingguistic books. In how many ways can be arrange the books, if books of the same topic must be placed together?


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NGUYỄN SANH KIÊN
27/03/2017 at 20:21
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3
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Given three number a,b,c know :

0 ≤≤ a ≤≤ b + 1 ≤≤ c + 2 and a + b + c = 1 .

Find the smallest value of c

  • ...
    MCP 27/03/2017 at 20:22

    We have : a ≤ ≤ c + 2 b + 1 ≤ ≤ c + 2 c + 2 = c + 2 => a + b + 1 + c + 2 ≤ ≤ c + 2 + c + 2 + c + 2 => (a + b + c) + (1 + 2) ≤ ≤ 3c + 6 => 1 + 3 ≤ ≤ 3c + 6 => 4 ≤ ≤ 3c + 6 => − 2 3 −23 ≤ ≤ c => Minc = − 2 3 −23 => a + b +( − 2 3 −23) = 1 => a + b = 5 3 53 With a = b + 1 => a + b = 5 3 53 => b + 1 + b = 5 3 53 => b = 1 3 13 => a = 4 3 43 (1) With a < b + 1 => b + 1 + b < 5 3 53 => b < 1 3 13 => a < 4 3 43 (2) From (1) and (2) => b ≤ 1 3 ≤13 and a ≤ 4 3 a≤43 So the smallest value of c = − 2 3 −23 When b ≤ 1 3 ≤13 and a ≤ 4 3

    NGUYỄN SANH KIÊN selected this answer.
  • ...
    NGUYỄN SANH KIÊN 29/03/2017 at 20:07

    cái đồ copy bài

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    Nếu bây giờ ngỏ ý . Liệu có còn kịp không 28/03/2017 at 12:35
    We have : a ≤ ≤ c + 2 b + 1 ≤ ≤ c + 2 c + 2 = c + 2 => a + b + 1 + c + 2 ≤ ≤ c + 2 + c + 2 + c + 2 => (a + b + c) + (1 + 2) ≤ ≤ 3c + 6 => 1 + 3 ≤ ≤ 3c + 6 => 4 ≤ ≤ 3c + 6 => − 2 3 −23 ≤ ≤ c => Minc = − 2 3 −23 => a + b +( − 2 3 −23) = 1 => a + b = 5 3 53 With a = b + 1 => a + b = 5 3 53 => b + 1 + b = 5 3 53 => b = 1 3 13 => a = 4 3 43 (1) With a < b + 1 => b + 1 + b < 5 3 53 => b < 1 3 13 => a < 4 3 43 (2) From (1) and (2) => b ≤ 1 3 ≤13 and a ≤ 4 3 a≤43 So the smallest value of c = − 2 3 −23 When b ≤ 1 3 ≤13 and a ≤ 4 3

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Phúc Hoàng
30/03/2017 at 21:51
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0
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Let a,b,c satisfy a^2+b^2+c^2=1 .Prove that abc+2(1+a+b+c+ab+bc+ca) \(\ge\) 0

help


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Nguyễn Đức Kiên
01/04/2017 at 12:27
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6
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Tom and Tony have all 100 candy.Find out the number of candy you each know that Tom's candy is 1/4 of Tony's candy.

  • ...
    Mina Phạm 01/04/2017 at 14:39

    ĐS: Tom 20 cái kẹo

    Tony:80 cái kẹo

    Nguyễn Đức Kiên selected this answer.
  • ...
    Vũ Việt Vương 01/04/2017 at 16:50

    Tom 20 candy

    Tony 80 candy

  • ...
    Nguyễn Phương Linh 08/04/2017 at 14:47

    The number of candy that Tom have is:

    100 : (1 + 4) x 1 = 20 (candies)

    The number of candy that Tony have is:

    100 - 20 = 80 (candies)

                   Answer: Tom: 20 candies.

                                  Tony: 80 candies.


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Nguyễn Nhật Minh
02/04/2017 at 22:06
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Prove that \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\left(n\in N;n\ge2\right)\)

Help me please!

Fractioninequality

  • ...
    Nguyễn Huy Tú 03/04/2017 at 13:03

    \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\)

    \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\)

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

    \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\)

    \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\)

    \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

    So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

    Nguyễn Nhật Minh selected this answer.
  • ...
    AI kết bạn với mình là may mắn cả đời 04/04/2017 at 21:05

    \( \dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\) \(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\) \(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

  • ...
    Indrace AD-MIN 07/04/2017 at 21:35

    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


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Phan Minh Anh
15/06/2017 at 12:55
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3
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Quick count:

50x100x1x25x4x2x5x20=?

  • ...
    Phan Minh Anh 15/06/2017 at 13:07

    50x100x1x25x4x2x5x20

    =(50x2)x(100x1)x(25x4)x(20x5)

    =  100   x   100    x   100  x  100

    =     10000           x      10000

    =      100000000.

  • ...
    Nguyễn Tiến Dũng 15/06/2017 at 15:21

    (50x2)x(100x1)x(20x5)x(25x4)

    =100x100x100x100

    =1004

  • ...
    Phan Thanh Tinh Coordinator 15/06/2017 at 13:07

    50 x 100 x 1 x 25 x 4 x 2 x 5 x 20

    = (50 x 2) x 100 x (25 x 4) x (5 x 20)

    = 100 x 100 x 100 x 100

    = (102)4 = 108


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Huyền Phạm
15/07/2017 at 15:37
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0
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2002 is a number that stays the same when read backwards as when read forwards. Which the flollowing does not have this propety?


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FA Thần Tốc Độ
16/01/2018 at 21:24
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3
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Polynomial Analysis into Factors :

a ,  \(x^8+14x^4+1\)

b , \(x^8+98x^4+1\)

 

  • ...
    FA FIFA Club World Cup 2018 16/01/2018 at 21:32

    a , \(x^8+14x^4+1=\left(x^4+1\right)^2+12x^4\)

    Add and subtract \(4x^2\left(x^4+1\right)\) we have :

    \(\left(x^4+1\right)^2+4x^2\left(x^4+1\right)+4x^4-4x^2\left(x^4+1\right)+8x^4\)

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

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

    \(=\left(x^4+2x^2+1\right)^2-\left(2x^3-2x\right)^2\)

    \(=\left(x^4+2x^3+2x^2-2x+1\right)\left(x^4-2x^3+2x^2+2x+1\right)\)

    FA Thần Tốc Độ selected this answer.
  • ...
    FA FIFA Club World Cup 2018 16/01/2018 at 21:39

    b , \(x^8+98x^4+1=\left(x^4+1\right)^2+96x^4\)

    \(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)

    \(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)\)

    \(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^2-1\right)^2\)

    \(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)

    \(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)

  • ...
    Fc Alan Walker 22/01/2018 at 12:29

    a , x8+14x4+1=(x4+1)2+12x4

    Add and subtract 4x2(x4+1)

    we have :

    (x4+1)2+4x2(x4+1)+4x4−4x2(x4+1)+8x4

    =(x4+1+2x2)2−4x2(x4+1−2x2)

    =(x4+2x2+1)2−4x2(x2−1)2

    =(x4+2x2+1)2−(2x3−2x)2

    =(x4+2x3+2x2−2x+1)(x4−2x3+2x2+2x+1)

    b , x8+98x4+1=(x4+1)2+96x4

    =(x4+1)2+16x2(x4+1)+64x4−16x2(x4+1)+32x4

    =(x4+1+8x2)2−16x2(x4+1−2x2)

    =(x4+8x2+1)2−16x2(x2−1)2

    =(x4+8x2+1)2−(4x3−4x)2

    =(x4+4x3+8x2−4x+1)(x4−4x3+8x2+4x+1)


...
DJ Shape of you
16/01/2018 at 21:42
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3
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Solve the equation :

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

  • ...
    FA FIFA Club World Cup 2018 16/01/2018 at 21:48

    We have : \(\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

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

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

    \(\dfrac{3x-2}{x-1}-\dfrac{1}{x-1}=0\)

    \(\dfrac{3x-2-1}{x-1}=0\)

    \(\dfrac{3x-3}{x-1}=0\)

    \(\dfrac{3\left(x-1\right)}{x-1}=0\)

    \(3=0\)

    So the equation has no solution .


     

    DJ Shape of you selected this answer.
  • ...
    Kaya Renger Coordinator 17/01/2018 at 21:04

    Condiction : x \(\ne\) 1

    Following FA FIFA's answer , at the sixth row , we have :

    \(\dfrac{3x-3}{x-1}=0\)

    \(\Leftrightarrow\dfrac{3\left(x-1\right)}{x-1}=0\)

    \(\Leftrightarrow3\left(x-1\right)=0\)

    <=> x = 1 

    Impossible because for equation exists , x must different to 1

    It also mean that there is not any roots satisfy equation .

    P/s : FIFA's answer was wrong because a few equation you cannot compact the fraction . In equation , you must to use this sign "<=>" to indicate an equivalent equation !

  • ...
    Fc Alan Walker 22/01/2018 at 12:28

    We have : 2x−1x−1+1=1x−1

    2x−1x−1+x−1x−1=1x−1

    2x−1+x−1x−1=1x−1

    3x−2x−1−1x−1=0

    3x−2−1x−1=0

    3x−3x−1=0

    3(x−1)x−1=0

    3=0

    So the equation has no solution .


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α β γ δ θ σ ∂ ε ω φ ϕ π μ λ Ψ ξ η χ ζ ι κ ν ψ Ω ρ τ υ Γ Δ Λ Φ Π Σ Υ Ξ ϑ Θ ς ϰ
∞ ⊻ ⩞ ⋎ ⋏ ≀ ∪ ⊎ ⋓ ∩ ⋒ ⊔ ⊓ ⨿ ⊗ ⊙ ⊚ ⊛ ⊘ ⊝ ⊕ ⊖ ⊠ ◯ ⊥
⇔ ⇒ ⇐ → ← ↔ ↑ ↓
Operations
+ - ÷ × ≠ = ⊂ ⊃ ⊆ ⊇ ≈ ∈ ∉ ∃ ∄ ≤ ≥ ± ∓ ≠ ∅ ≃ ≅ ≡ ⋮ ⋮̸ ∀
(□) [□] {□} |□|

The type of system

m×n 1×2 1×3 1×4 1×5 1×6
2×1 2×2 2×3 2×4 2×5 2×6
3×1 3×2 3×3 3×4 3×5 3×6
4×1 4×2 4×3 4×4 4×5 4×6
5×1 5×2 5×3 5×4 5×5 5×6
6×1 6×2 6×3 6×4 6×5 6×6

Recipe:

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