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absolute value

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Cristiano Ronaldo
19/03/2017 at 10:42
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If m + |m| + n = 8 and  |n| + m-n = 9, find m-n.

absolute value


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Cristiano Ronaldo
19/03/2017 at 10:38
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Find the smallest value of |a-1000| + |a-1001|.

absolute value

  • ...
    ¤« 08/04/2018 at 15:11

    We have:

    |a−1001|≥0

    ⇒|a−1000|≥1

    ⇒|a−1000|+|a−1001|≥1+0=1

    => Smallest value of |a-1000| + |a - 1001| = 1 at:

    * a = 1001 and |a-1000| + |a - 1001| = 1 + 0 = 1

    * a = 1000 and |a-1000| + |a - 1001| = 0 + 1 = 1

    Are you OK?

  • ...
    Dao Trong Luan 24/07/2017 at 11:51

    We have:

    \(\left|a-1001\right|\ge0\)

    \(\Rightarrow\left|a-1000\right|\ge1\)

    \(\Rightarrow\left|a-1000\right|+\left|a-1001\right|\ge1+0=1\)

    => Smallest value of |a-1000| + |a - 1001| = 1 at:

    * a = 1001 and |a-1000| + |a - 1001| = 1 + 0 = 1

    * a = 1000 and |a-1000| + |a - 1001| = 0 + 1 = 1

    Are you OK?

  • ...
    Fujitora Ishito (online math) 20/03/2017 at 15:56

    smallest value : 1

    When a=1000 or a=1001 

    Good


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Cristiano Ronaldo
19/03/2017 at 10:37
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Simplify |a+1| + |a-1| for -1 <= a <=0.

absolute value

  • ...
    ¤« 08/04/2018 at 15:11

    −1≤a≤0⇒{0≤a+1a−1<0⇒{|a+1|=a+1|a−1|=1−a

    => |a + 1| + |a - 1| = a + 1 + 1 - a = 2

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 18:16

    \(-1\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+1\\a-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+1\right|=a+1\\\left|a-1\right|=1-a\end{matrix}\right.\)

    => |a + 1| + |a - 1| = a + 1 + 1 - a = 2


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Jeff Bezos
19/03/2017 at 10:35
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The values of a,b and c are depicted on them number line.

| | | | > c b 0 a

Simplify |a+b| + |a-b| + |c-a|

absolute value

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 20:19

    Can I ask that the distance from b to 0 is longer than the distance from 0 to a ?


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Jeff Bezos
19/03/2017 at 10:29
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What will be the smallest value for |x-1| + |x+2| + |x+3|?

absolute value


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Jeff Bezos
19/03/2017 at 10:28
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Find the values of x for ||4x-2| - 2| = 4

absolute value

  • ...
    ¤« 08/04/2018 at 15:12

    ||4x-2| - 2| = 4

    Scenerio (S.c.) 1:

    |4x - 2| - 2 = 4

    |4x - 2| = 6

    S.c. 1.1:

    4x - 2 = 6

    4x = 8

    x = 2.

    S.c. 1.2:

    4x - 2 = -6

    4x = -4

    x = -1.

    S.c. 2:

    |4x - 2| - 2 = -4

    |4x - 2| = -2

    |4x - 2| is always larger or equal to 0, so there is no value of x in this scenerio.

    Thus, the values of x are 2; -1.

  • ...
    Nguyễn Nhật Minh 21/03/2017 at 18:16

    ||4x-2| - 2| = 4

    Scenerio (S.c.) 1:

    |4x - 2| - 2 = 4

    |4x - 2| = 6

    S.c. 1.1:

    4x - 2 = 6

    4x = 8

    x = 2.

    S.c. 1.2:

    4x - 2 = -6

    4x = -4

    x = -1.

    S.c. 2:

    |4x - 2| - 2 = -4

    |4x - 2| = -2

    |4x - 2| is always larger or equal to 0, so there is no value of x in this scenerio.

    Thus, the values of x are 2; -1.


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Jeff Bezos
19/03/2017 at 10:27
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Find the values of x for |2x-3| + |2x+1| = 4

absolute value

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    Phan Thanh Tinh Coordinator 23/03/2017 at 20:31

    Case 1 :\(x< -\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}2x-3< -4< 0\\2x+1< 0\end{matrix}\right.\)

    \(\Rightarrow3-2x-2x-1=4\Rightarrow4x=-2\Rightarrow x=-\dfrac{1}{2}\)(asburd)

    Case 2 :\(-\dfrac{1}{2}\le x< \dfrac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3< 0\\2x+1\ge0\end{matrix}\right.\)

    \(\Rightarrow3-2x+2x+1=4\Rightarrow4=4\) (true)

    Case 3 :\(x\ge\dfrac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3\ge0\\2x+1\ge4>0\end{matrix}\right.\)

    \(\Rightarrow2x-3+2x+1=4\Rightarrow4x-2=4\Rightarrow4x=6\Rightarrow x=\dfrac{3}{2}\)(true)

    So\(-\dfrac{1}{2}\le x\le\dfrac{3}{2}\)

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:12

    Case 1 :x<−12⇔{2x−3<−4<02x+1<0

    ⇒3−2x−2x−1=4⇒4x=−2⇒x=−12

    (asburd)

    Case 2 :−12≤x<32⇒{2x−3<02x+1≥0

    ⇒3−2x+2x+1=4⇒4=4

     (true)

    Case 3 :x≥32⇒{2x−3≥02x+1≥4>0

    ⇒2x−3+2x+1=4⇒4x−2=4⇒4x=6⇒x=32

    (true)

    So−12≤x≤32


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Jeff Bezos
19/03/2017 at 10:27
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If a - b > b + a then

absolute value

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    Phan Thanh Tinh Coordinator 23/05/2017 at 09:16

    \(a-b>b+a\Rightarrow a-b-b-a>b+a-b-a\)

    \(\Rightarrow-2b>0\Rightarrow b< 0\)

  • ...
    Fujitora Ishito (online math) 20/03/2017 at 15:58

    b = \(^{Z^-}\)

    Good ^^ 


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Jeff Bezos
19/03/2017 at 10:20
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What is the range of values of x for |x-3| + (x+3) = 0?

absolute value

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    Phan Thanh Tinh Coordinator 23/03/2017 at 20:18

    \(\left|x-3\right|+\left(x+3\right)=0\)

    Case 1 :\(x-3\ge0\Leftrightarrow x\ge3\),so we have :

    x - 3 + x + 3 = 0 => 2x = 0 => x = 0 (doesn't satisfy \(x\ge3\))

    Case 2 :\(x-3< 0\Leftrightarrow x< 3\),so we have :

    3 - x + x + 3 = 0 => 6 = 0 (absurd)

    So there are no values of x

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:12

    |x−3|+(x+3)=0

    Case 1 :x−3≥0⇔x≥3

    ,so we have :

    x - 3 + x + 3 = 0 => 2x = 0 => x = 0 (doesn't satisfy x≥3

    )

    Case 2 :x−3<0⇔x<3

    ,so we have :

    3 - x + x + 3 = 0 => 6 = 0 (absurd)

    So there are no values of x


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Jeff Bezos
19/03/2017 at 10:19
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Given that -2 < a < -1 amd 0 < b < 1, how many of the statements below has/have negative values ?

(a) |a+b|

(b) b-2a

(c) |b| - |a|

(d) |a+2|

(e) -|b-4|

absolute value


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Jeff Bezos
19/03/2017 at 10:17
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Which of the follwoing is true if a, b, c satisfy c < b <0 and 1 < a ?

absolute value


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Jeff Bezos
19/03/2017 at 01:02
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If a < 0, the value of 3a + 8|a| is ?

absolute value

  • ...
    Gấu Baby 19/03/2017 at 08:52

    3a + 8|a| = (-3)|a|+8|a|

    =\([(-3)+8]\)|a|

    =5|a|

    Jeff Bezos selected this answer.

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steve jobs
17/03/2017 at 11:00
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The corresponding values of a,b and c are shown in the figure below.

Simplify |b - a| + |b + c| - |c| for a > b > c.

> | | | | c b 0 a

absolute value

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 19:15

    We have :

    \(\left\{{}\begin{matrix}b< a\\b< 0\\c< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b-a< 0\\b+c< 0\\c< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left|b-a\right|=a-b\\\left|b+c\right|=-b-c\\\left|c\right|=-c\end{matrix}\right.\)

    => |b - a| + |b + c| + |c| = a - b - b - c - c = a - 2b - 2c

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:12

    We have :

    ⎧⎩⎨⎪⎪b<ab<0c<0⇒⎧⎩⎨⎪⎪b−a<0b+c<0c<0

    ⇒⎧⎩⎨⎪⎪|b−a|=a−b|b+c|=−b−c|c|=−c

    => |b - a| + |b + c| + |c| = a - b - b - c - c = a - 2b - 2c


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steve jobs
17/03/2017 at 10:51
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Find the smallest value of |m - 2| + |m + 3|.

absolute value

  • ...
    Nguyễn Huy Tú 17/03/2017 at 12:55

    Put \(A=\left|m-2\right|+\left|m+3\right|\)

    We have: \(A=\left|m-2\right|+\left|m+3\right|=\left|2-m\right|+\left|m+3\right|\)

    Apply the inequality \(\left|a\right|+\left|b\right|\ge\left|a\right|+\left|b\right|\), we have:

    \(A\ge\left|2-m+m+3\right|=\left|5\right|=5\)

    The "=" sign occurs when \(\left\{{}\begin{matrix}2-m\ge0\\m+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m\le2\\m\ge-3\end{matrix}\right.\Rightarrow-3\le m\le2\)

    So \(MIN_A=5\) when \(-3\le m\le2\)

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:13

    Put A=|m−2|+|m+3|

    We have: A=|m−2|+|m+3|=|2−m|+|m+3|

    Apply the inequality |a|+|b|≥|a|+|b|

    , we have:

    A≥|2−m+m+3|=|5|=5

    The "=" sign occurs when {2−m≥0m+3≥0⇒{m≤2m≥−3⇒−3≤m≤2

    So MINA=5

     when −3≤m≤2

    Find minimize |m−2|+|m+3|

     ?

    By inequality |a|+|b|≥|a+b|

     we have:

    |m−2|+|m+3|=|2−m|+|m+3|

    ≥|2−m+m+3|=5

    Done !

  • ...
    Ace Legona 17/03/2017 at 11:52

    Find minimize \(|m-2|+|m+3|\) ?

    By inequality \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) we have:

    \(\left|m-2\right|+\left|m+3\right|=\left|2-m\right|+\left|m+3\right|\)

    \(\ge\left|2-m+m+3\right|=5\)

    Done !


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Nguyễn Đức Mạnh
17/03/2017 at 10:49
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Find the value of |b - a + 1| - |a - b -3| for a < 0 and ab < 0.

absolute value

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 19:30

    \(\left\{{}\begin{matrix}a< 0\\ab< 0\end{matrix}\right.\Rightarrow b>0\Rightarrow b>a\Rightarrow\left\{{}\begin{matrix}b-a+1>1>0\\a-b-3< -3< 0\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}\left|b-a+1\right|=b-a+1\\\left|a-b-3\right|=-\left(a-b-3\right)\end{matrix}\right.\)

    \(\Rightarrow\left|b-a+1\right|-\left|a-b-3\right|=b-a+1+a-b-3=-2\)

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:13

    {a<0ab<0⇒b>0⇒b>a⇒{b−a+1>1>0a−b−3<−3<0

    ⇒{|b−a+1|=b−a+1|a−b−3|=−(a−b−3)

    ⇒|b−a+1|−|a−b−3|=b−a+1+a−b−3=−2


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Nguyễn Đức Mạnh
17/03/2017 at 10:48
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Simplify |a+1| + |a-1| for    -1 <= a <= 0 

absolute value

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 19:43

    \(-1\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+1\\a-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+1\right|=a+1\\\left|a-1\right|=-\left(a-1\right)\end{matrix}\right.\)

    \(\Rightarrow\left|a+1\right|+\left|a-1\right|=a+1-\left(a-1\right)=2\)

    Selected by MathYouLike
  • ...
    ¤« 08/04/2018 at 15:14

    −1≤a≤0⇒{0≤a+1a−1<0⇒{|a+1|=a+1|a−1|=−(a−1)

    ⇒|a+1|+|a−1|=a+1−(a−1)=2


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Nguyễn Đức Mạnh
17/03/2017 at 10:47
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It is given |a| = 7, |b| = 5 and |a - b| = b - a.

Find the value of a + b

absolute value

  • ...
    ¤« 08/04/2018 at 15:14

    |a|=7;|b|=5⇒(a;b)=(7;5);(−7;5);(7;−5);(−7;−5)

    We have :|a−b|=b−a⇒a−b≤0⇒a≤b

    So (a ; b) = (-7 ; 5) ; (-7 ; -5)

    => a + b = -2 or a + b = -12

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 19:58

    \(\left|a\right|=7;\left|b\right|=5\Rightarrow\left(a;b\right)=\left(7;5\right);\left(-7;5\right);\left(7;-5\right);\left(-7;-5\right)\) 

    We have :\(\left|a-b\right|=b-a\Rightarrow a-b\le0\Rightarrow a\le b\)

    So (a ; b) = (-7 ; 5) ; (-7 ; -5)

    => a + b = -2 or a + b = -12


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Nguyễn Đức Mạnh
17/03/2017 at 10:46
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Simplify |a+3| + |a-3| if the value of a is given as   -3 <= a <= 0.

absolute value

  • ...
    ¤« 08/04/2018 at 15:15

    −3≤a≤0⇒{0≤a+3a−3<0⇒{|a+3|=a+3|a−3|=−(a−3)

    ⇒|a+3|+|a−3|=a+3−(a−3)=6

  • ...
    Phan Thanh Tinh Coordinator 23/03/2017 at 19:45

    \(-3\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+3\\a-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+3\right|=a+3\\\left|a-3\right|=-\left(a-3\right)\end{matrix}\right.\)

    \(\Rightarrow\left|a+3\right|+\left|a-3\right|=a+3-\left(a-3\right)=6\)


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Nguyễn Đức Mạnh
17/03/2017 at 10:45
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It is given |a| = 2, |b| = 3, |c| = 4, a > b >c. Find the value of a + b -c.

absolute value

  • ...
    Đại Việt 17/03/2017 at 21:02

    Call A=a+b-c

    We have |a|=2 \(\Rightarrow\) a = 2 or a = (-2)

    |b|= 3 \(\Rightarrow\) b = 3 or b = (-3)

    |c|= 4 \(\Rightarrow\) c = 4 or c = (-4)

    \(\Rightarrow\)a = 2; b = 3;c = 4 or a = -2;b = -3;c = -4

    But a > b > c ( theme for )
    So a = -2;b = -3;c = -4 ( because -2 > -3 > -4 )

    Substitute a = -2;b = -3;c = -4 to A, we have :

    A= -2+(-3)-(-4)

    \(\Rightarrow\)A= -2-3+4

    \(\Rightarrow\)A= -5+4

    \(\Rightarrow\)A= -1 

    \(\Rightarrow\) a+b-c= -1 in a = -2;b = -3;c = -4

     
     

  • ...
    ¤« 08/04/2018 at 15:15

    Call A=a+b-c

    We have |a|=2 ⇒

     a = 2 or a = (-2)

    |b|= 3 ⇒

     b = 3 or b = (-3)

    |c|= 4 ⇒

     c = 4 or c = (-4)

    ⇒

    a = 2; b = 3;c = 4 or a = -2;b = -3;c = -4

    But a > b > c ( theme for )
    So a = -2;b = -3;c = -4 ( because -2 > -3 > -4 )

    Substitute a = -2;b = -3;c = -4 to A, we have :

    A= -2+(-3)-(-4)

    ⇒

    A= -2-3+4

    ⇒

    A= -5+4

    ⇒

    A= -1 

    ⇒

     a+b-c= -1 in a = -2;b = -3;c = -4

     
     

  • ...
    Nguyễn Huy Tú 17/03/2017 at 13:06

    We have: \(\left|a\right|=2\Rightarrow a=2\) or a = -2

    \(\left|b\right|=3\Rightarrow b=3\) or b = -3

    \(\left|c\right|=4\Rightarrow c=4\) or c = -4

    \(\Rightarrow\left(a;b;c\right)\in\left(2;3;4\right);\left(-2;-3;-4\right)\)

    but \(a>b>c\)

    \(\Rightarrow\left(a;b;c\right)\in\left(-2;-3;-4\right)\)

    \(\Rightarrow a+b-c=-2-3+4=-1\)

    So the value of a + b - c = -1


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