{a+b=3a2+b2≥5⇒{a2+2ab+b2=9a4+2a2b2+b4≥25⇒{2ab=9−(a2+b2)a4+b4+2a2b2≥25

We have : a2 + b2 ≥

 5

=> −(a2+b2)≤5

=> 9−(a2+b2)≥9−5=4

=> 2ab≥4

=> ab≥2

<=> a2b2≥4

<=> 4a2b2≥16

Plus 4a2b2≥16

 into a4+b4+2a2b2≥25

=> a4+b4+6a2b2≥41

=> Min = 41

That is my opinion :v

There is something is wrong here 

Change to :

We have a2+b2≥5

=> 9−(a2+b2)≤4

=> 2ab≤4

=> ab≤2

<=> a2b2≥4

=> 4a2b2≥16

Jim has the number of apples:

20 : 5 x 4 = 16 (apples)

Answer 16 apples

Wrong post

Because if a,b,c > 0 so

⎧⎩⎨⎪⎪a≥a−−√b≥b√c≥c√⇒a+b+c≥a−−√+b√+c√

⎪mn=pnp=mmp=n⇒mn.np.mp=p.m.n⇔m2.n2.p2=m.n.p

⇒m.n.p.(m.n.p−1)=0

⇔[m.n.p=0m.n.p=1

With m.n.p=0⇒p2=0⇒p=0⇒m=n=0

With m.n.p=1⇒p2=1⇒p=±1

  If p=1

 then {mn=1m=n⇒[m=n=1m=n=−1

  If  p=−1

 then {mn=−1m=−n⇒[m=1,n=−1m=−1,n=1

 So have 5 ordered triples of integers satisfied

9 bills of 100 euro: 100 x 9 = 900 ( euro )

9 bills of 10 euro: 10 x 9 = 90 ( euro )

10 coins of 1 euro: 10 x 1 = 10 ( euro )

He has: 900 + 90 + 10 = 1000 ( euro )

            Answer: 1000 euro

AB = AC = 8 => ΔABC

 is a isosceles triangle

=> AM is high road, median (trung tuyến) of this triangle.

=> AM ⊥

 BC and MB = MC = 5

=> AM = 82−52−−−−−−√=39−−√

a, f(2)=23−4.22−7.2−1=8−16−14−1=−23

g(12)=(12)4+(12)3−(12)2−7.12+2=116+18−14−72+2=−2516

b,

f(x)−g(x)=x3−4x2−7x−1−(x4+x3−x2−7x+2)

=x3−4x2−7x−1−x4−x3+x2+7x−2

=−x4−3x2−3

Jim tall is cm :

139 + 12 = 151 (cm)

Bob tall is cm:

151 - 21 = 130 ( cm)

Answer

Tell the pencil is a, the paper clips is b and the eraser is c

=> a + 5b = 2c and a = 29b

=> 29b + 5b = 2c

=> 34b = 2c

=> c = 17b

So 17 paper clips weigh the same as an eraser.

We have:y+z+1x=x+z+2y=x+y−3z=1x+y+z

=y+z+1+x+z+2+x+y−3x+y+z=2x+2y+2zx+y+z=2

⇒⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪y+z+1=2x(⋅)x+z+2=2y(⋅⋅)x+y−3=2z(⋅⋅⋅)x+y+z=12(⋅⋅⋅⋅)

(⋅⋅⋅⋅)−(⋅)⇒x−1=12−2x⇒3x=32⇒x=12

(⋅⋅⋅⋅)−(⋅⋅)⇒y−2=12−2y⇒3y=52⇒y=56

(⋅⋅⋅⋅)−(⋅⋅⋅)⇒z+3=12−2z⇒3z=−52⇒z=−56

⇒P=2016x+y2017+z2017=2016.12+(56)2017+(−56)2017

=1008+(56)2017−(56)2017=1008

Sorry for OLD answer

First ask a simpler question: What is the maximum number of points of intersection for a line and a square? Because the square is convex, the answer is 2.

A Triangle consists of 3 line segments, each of which is a subset of a line. So each segment can clearly not intersect the square more than twice. This gives 6 as the upper bound, even if you extend each edge of the triangle into an infinite line.

Further note that if both endpoints of an edge are inside the square, then the whole edge is inside the square and doesn’t intersect it at all. And if one endpoint is inside and one is outside, then there can only be one point of intersection.

Therefore, you can only get 6 points of intersection if all 3 corners of the triangle are outside the square. If one is inside, the max is 4; if two are inside, the max is 2; and if 3 are inside, the max is 0.

We have:Vhinhcau=3Shinhcau

⇒43.π.r3=3.4.π.r2

 with r

 is the radius of this certain

⇒r3r2=3.4.π43.π=3.443=3.4.34=9

⇒r=9

Call Amos is A, Butch is B and Cody is 3

C1:  - If A is the one to shoot first and shoot C, he will be killed by B

     => SO A will shoot B first

     Then it's time for C to shoot A

     => The percentage of A alive is 50% and C is always alive (out of shot)

 C2: - If B is the one to shoot first and shoot C, he will be killed by A

     => SO B will shoot A first

     Then it's time for C to shoot B

     => The percentage of B alive is 50% and C is always alive (out of shot)

 - If C is the one to shoot first 

     + He shoots A:

          C3. If shoot to target, B will kill C => B is alive

          C4. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)

  + He shoots B:

          C5. If shoot to target, A will kill C

          C6. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)

SO WE HAVE:

          ALIVE        50%        DEAD

C1:         C               A               B

C2:         C               B               A

C3:           B                              A , C

C4:       C, A (or B)                B (or A)

C5:         A                               B, C             

C6:       C, A (or B)                B (or A)         

SO C HAS THE BIGGEST CHANGE TO LIVE (46=66.66%

 )

     A (or B)' s probability to live;1−462=16=16.66%

123-45-67+89=100

1 + 2 + 3 – 4 + 5 + 6 + 78 + 9 = 100
123 +45 - 67 + 8 - 9 = 100

123 + 4- 5 + 67 - 89 = 100
 

We have:1−11+2+.......+n=1−1n(n+1)2=1−2n(n+1)

=n(n+1)−2n(n+1)=n2+n−2n(n+1)=n2−n+2n−2n(n+1)=n(n−1)+2(n−1)n(n+1)=(n+2)(n−1)n(n+1)

Apply we have:(1−11+2)(1−11+2+3)..........(1−11+2+.......+2006)

=4.12.3.5.23.4........2008.20052006.2007=4.5.....20082.3.....2006.1.2......20053.4......2007=2007.20082.3.1.22006.2007

=20083.2006=10043.1003=10043009

16) We have: 

SOLUTION 1

8 + 1 + 3 = 12; 1 + 2 = 3

9 + 0 + 9 = 18; 1 + 8 = 9

4 + 5 + 3 = 12; 1 + 2 = 3

So 1 + 4 + 6 + 2 = 13; 1 + 3 = 4.

SOLUTION 2

8 + 1 + 3 = 12; 12 - 9 = 3

9 + 0 + 9 = 18; 18 - 9 = 9

4 + 5 + 3 = 12; 12 - 9 = 3

So 1 + 4 + 6 + 2 = 13; 13 - 9 = 4

ANSWER: 4

17) G=34+536+7144+...+2n+1n2(n+1)2

G=1−122+122−132+...+1n2−1(n+1)2

G=1−1(n+1)2<1

 (proved)

Draw the equilateral triangle EBC

Consider Δ

AEB and Δ

AEC, we have:

- AE is common edge (hypothesis) 

- AB = AC (hypothesis) 

- EB = EC (Δ

EBC is a equilateral triangle) 

=> Δ

AEB = Δ

AEC (S-S-S) 

=> EABˆ=EACˆ=20o2=10o

Δ

ABC balance at A => ABCˆ=ACBˆ=180o−20o2=80o

Because Δ

EBC is a equilateral triangle so EBCˆ=ECBˆ=60o

⇒EBAˆ=ECAˆ=80o−60o=20o

Consider Δ

AEC and Δ

CDA, we have:

- AC is common edge (hypothesis) 

- DACˆ=ACEˆ=20o

- AD = EC (= BC) 

=> Δ

AEC = Δ

CDA (S-A-S) 

=> ACDˆ=EACˆ=10o

(3x+2)(3x+3)(3x+4)=(3x+5−3)(3x+5−2)(3x+5−1)=(13−3)(13−2)(13−1)=10.11.12=1320

Let x, y be the length and breadth of a rectangle and 16 is the length of diagonal as shown below:

Using the Pythagorean Theorem in the right triangle formed by sides with the diagonal, we get: x2 + y2 = 2562

=> y=256−x2−−−−−−−√

 and A(x)=x256−x2−−−−−−−√

that is the function we have to maximize. Notice that x can vary between x = 0 (the rectangle collapses to a vertical line) and x = 16 (the rectangle collapses to a horizontal line), and in both extremes the area is 0. Let’s find critical points for A:

0=A′(x)=256−x2−−−−−−−√−x−x256−x2−−−−−−−√⇒x2256−x2−−−−−−−√=16−x2−−−−−−√⇒x2=256−x2

and so x=128−−−√

 is the only critical value

Moreover, A(8–√

) = 8 > 8 so it is the maximum

x y 16