absolute value
Cristiano Ronaldo
19/03/2017 at 10:42
Cristiano Ronaldo
19/03/2017 at 10:38
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¤« 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?
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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?
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¤« 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
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\(-1\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+1\\a-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+1\right|=a+1\\\left|a-1\right|=1-a\end{matrix}\right.\)
=> |a + 1| + |a - 1| = a + 1 + 1 - a = 2
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Can I ask that the distance from b to 0 is longer than the distance from 0 to a ?
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¤« 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.
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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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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\(-\dfrac{1}{2}\le x\le\dfrac{3}{2}\)
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¤« 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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\(a-b>b+a\Rightarrow a-b-b-a>b+a-b-a\)
\(\Rightarrow-2b>0\Rightarrow b< 0\)
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Jeff Bezos
19/03/2017 at 10:20
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\(\left|x-3\right|+\left(x+3\right)=0\)
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
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¤« 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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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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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
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¤« 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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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 \(-3\le m\le2\)
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¤« 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 !
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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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\(\left\{{}\begin{matrix}a< 0\\ab< 0\end{matrix}\right.\Rightarrow b>0\Rightarrow b>a\Rightarrow\left\{{}\begin{matrix}b-a+1>1>0\\a-b-3< -3< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|b-a+1\right|=b-a+1\\\left|a-b-3\right|=-\left(a-b-3\right)\end{matrix}\right.\)
\(\Rightarrow\left|b-a+1\right|-\left|a-b-3\right|=b-a+1+a-b-3=-2\)
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¤« 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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\(-1\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+1\\a-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+1\right|=a+1\\\left|a-1\right|=-\left(a-1\right)\end{matrix}\right.\)
\(\Rightarrow\left|a+1\right|+\left|a-1\right|=a+1-\left(a-1\right)=2\)
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¤« 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
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\(\left|a\right|=7;\left|b\right|=5\Rightarrow\left(a;b\right)=\left(7;5\right);\left(-7;5\right);\left(7;-5\right);\left(-7;-5\right)\)
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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\(-3\le a\le0\Rightarrow\left\{{}\begin{matrix}0\le a+3\\a-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|a+3\right|=a+3\\\left|a-3\right|=-\left(a-3\right)\end{matrix}\right.\)
\(\Rightarrow\left|a+3\right|+\left|a-3\right|=a+3-\left(a-3\right)=6\)
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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
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¤« 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
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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