Voted Questions
Asuna Yuuki
09/03/2017 at 21:33-
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3000cm3 = 30 litter
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FA FIFA Club World Cup 2018 16/01/2018 at 22:00
3000cm3 = 3 litter
NGUYỄN SANH KIÊN
27/03/2017 at 20:21-
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 12:35We 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
Phúc Hoàng
30/03/2017 at 21:51Nguyễn Đức Kiên
01/04/2017 at 12:27-
Mina Phạm 01/04/2017 at 14:39
ĐS: Tom 20 cái kẹo
Tony:80 cái kẹo
Nguyễn Đức Kiên selected this answer. -
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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.
Nguyễn Nhật Minh
02/04/2017 at 22:06-
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 \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)
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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
Phan Minh Anh
15/06/2017 at 12:55-
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.
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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
Huyền Phạm
15/07/2017 at 15:37FA Thần Tốc Độ
16/01/2018 at 21:24-
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 , \(x^8+98x^4+1=\left(x^4+1\right)^2+96x^4\)
\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^2-1\right)^2\)
\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)
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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-
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.
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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 !
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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 .