We have:

CD.DA = 24; FD⋅DA2=9⇒FD⋅DA=18

=> FDCD=1824=34⇒FC=14CD

AB⋅BE2=4⇒AB⋅BE=8

⇒BEBC=824=13⇒EC=23BC

⇒SΔEFC=14CD⋅23BC2=16.242=42=2(cm2)

⇒SΔAEF=24−9−4−2=13(cm2)

202 + ⊠⊠⊠ + ⊠⊠⊠ = 2002

=> ⊠⊠⊠ + ⊠⊠⊠ = 2002 - 202 = 1800

We have: 819 + 981 = 1800; 849 + 951 = 1800;.........

So the sum of all the 6 missing digits are: 8 + 1 + 9 + 9 + 8 + 1 = 8 + 4 + 9 + 9 + 5 + 1 = ... = 36

(Solution 1)We have: 1+1⋅2⋅3⋅4−−−−−−−−−−−√=1+1⋅4=5

1+2⋅3⋅4⋅5−−−−−−−−−−−√=1+2⋅5=11

.........

1+204⋅205⋅206⋅207−−−−−−−−−−−−−−−−−−√=1+204⋅207=42229

(Solution 2) We have: 1+1⋅2⋅3⋅4−−−−−−−−−−−√=2⋅3−1=5

1+2⋅3⋅4⋅5−−−−−−−−−−−√=3⋅4−1=11

..............

1+204⋅205⋅206⋅207−−−−−−−−−−−−−−−−−−√=205⋅206−1=42229

Alone has done solutions 3+4.

Solution 1:1+204.205.206.207−−−−−−−−−−−−−−−−√=1783288441−−−−−−−−−√=42229

Solution 2:1+n(n+1)(n+2)(n+3)−−−−−−−−−−−−−−−−−−−−−−√=1+(n2+3n)(n2+3n+2)−−−−−−−−−−−−−−−−−−−−−−√=(n2+3n)2+2.(n2+3n)+1−−−−−−−−−−−−−−−−−−−−−−−√

=(n2+3n+1)2−−−−−−−−−−−−√=n2+3n+1

With n=204 then 1+204.205.206.207−−−−−−−−−−−−−−−−√=2042+3.204+1=42229

A=(x−1)(x+6)(x+2)(x+3)=(x2+5x−6)(x2+5x+6)

=(x2+5x)2−62=(x2+5x)2−36≥−36

Then the minimum of A=-36

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

−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

|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

−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<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

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 !

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

|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

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

||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.

−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

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?

Change: 30%=310

Linda kept: 1−(15+310)=12

 of the 100 dollars as cash.

We have: S=(1101+1102+...+1150)+(1151+1152+...+1200)

                  >(1150+1150+...+1150)+(1200+1200+...+1200)

                                        50                                                      50

                  =50150+50200=13+14=712

So S>712

.

We have: A=(x+y)(x+2y)(x+3y)(x+4y)+y4

                    =[(x+y)(x+4y)][(x+2y)(x+3y)]+y4

                    =(x2+5xy+4y2)(x2+5xy+6y2)+y4

                    =[(x2+5xy+5y2)−y2][(x2+5xy+5y2)+y2]+y4

                    =(x2+5xy+5y2)2−y4+y4

                    =(x2+5xy+5y2)2

So A is a square number.

We can easily prove that P>0.

We have: P=122+132+...+120132<11.2+12.3+...+12012.2013

                                                                 =1−12013<1

⇒0<P<1

So P isn't a integer.