3000cm³ = 3 litter

3000cm³ = 30 litter

3000cm3 = 3 litter

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

cái đồ copy bài

ĐS: Tom 20 cái kẹo

Tony:80 cái kẹo

Tom 20 candy

Tony 80 candy

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.

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

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

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

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

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

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

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

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

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

50x100x1x25x4x2x5x20

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

=  100   x   100    x   100  x  100

=     10000           x      10000

=      100000000.

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

=100x100x100x100

=100⁴

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

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 :

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

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

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

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

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

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

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

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

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

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

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

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)

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 .

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 !

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 .