absolute value

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?

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?

smallest value : 1

When a=1000 or a=1001 

Good

−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

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

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

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

||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.

||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.

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

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

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

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

b = \(^{Z^-}\)

Good ^^ 

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

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

|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

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

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

=5|a|

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

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

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 \(-3\le m\le2\)

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 !

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 !

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

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

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

{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

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

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

−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

|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

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

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

−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

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

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

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

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

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