\(\Rightarrow\left[{}\begin{matrix}x+1+x+2+x+3+x+4=5x\\-x-1-x-2-x-3-x-4=5x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x+10=5x\\-4x-10=5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=-\dfrac{10}{9}\end{matrix}\right.\)

We have:

\(\left\{{}\begin{matrix}AB^2+AC^2=BC^2\\BH^2+AH^2=AB^2\\AH^2+HC^2=AC^2\end{matrix}\right.\Rightarrow BH^2+AH^2+AH^2+HC^2=BC^2\)

Because AB² + AC² = BC²

\(\Rightarrow BH^2+2\cdot AH^2+HC^2=BC^2\)

\(\Rightarrow9^2cm+2\cdot AH^2+4^2cm=13^2cm\)

\(\Rightarrow2\cdot AH^2=72cm\)

\(\Rightarrow AH^2=36cm^2\Leftrightarrow AH=6cm\)

So the area of triangle ABC is:

\(\dfrac{\left(9+4\right)\cdot6}{2}=\dfrac{13\cdot6}{2}=13\cdot3=39\left(cm^2\right)\) 

Because \(\overline{6x89y}⋮5\)

\(\Rightarrow\left[{}\begin{matrix}y=0\\y=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\overline{6x89y}=\overline{6x890}\\\overline{6x89y}=\overline{6x895}\end{matrix}\right.\)

And this number divisible by 9

\(\Rightarrow\left[{}\begin{matrix}6+x+8+9+0⋮9\\6+x+8+9+5⋮9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}23+x⋮9\\28+x⋮9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)

So \(\left(x,y\right)=\left(4;0\right);\left(8;5\right)\)

\(A=-2x^2+4x+8\)

\(=2\left(-x^2+2x+4\right)\)

\(=2\left[-\left(x^2-2\cdot x\cdot1+1\right)+5\right]\)

\(=2\left[-\left(x-1\right)^2+5\right]\)

\(=2\left[-\left(x-1\right)^2\right]+10\)

\(-\left(x-1\right)^2\le0\Rightarrow2\left[-\left(x-1\right)^2\right]\le0\)

\(\Rightarrow A\le10\)

\(\Rightarrow Max_A=10\) at x = 1

We have: 5^x = ...5 with \(\forall x\in\) N*

=> A = ...5 + ...5 + ... + ...5

And the number of terms of A are:

100 - 1 + 1 = 100 [terms]

So last-digit of A is: ...5 * 100 = ...500 = ...0

So, the answer is 0

Call Anne's teacher's age number is a

\(\Rightarrow\left\{{}\begin{matrix}20< a< 65\\a=x^2\end{matrix}\right.\Rightarrow a\in\left\{25;36;49;64\right\}\)

Because a > 2, if a+1 is a prime number

=> a+1 is a odd

=> a is a even number

\(\Rightarrow a\in\left\{36;64\right\}\)

** If a = 36 => a+1 = 37 --> Right a prime number 

** If a = 64 => a+1 = 65 divisive by 5 --> Wrong 

So teacher's Anne's age is 36 

a,

\(x^4+4\)

\(=\left[\left(x^2\right)^2+4x^2+4\right]-4x^2\)

\(=\left[\left(x^2\right)^2+2\cdot x^2\cdot2+2^2\right]-\left(2x\right)^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)   Apply \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)

b, 

\(4x^2+81\)

\(=2^2\cdot\left(x^2\right)^2+81\)

\(=\left[\left(2x^2\right)^2+36x^2+81\right]-36x^2\)

\(=\left[\left(2x^2\right)^2+2\cdot2x^2\cdot9+9^2\right]-\left(6x\right)^2\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)\)

I'm in 7th grade so I can not do sentence c

I know Phan Thanh Tinh, but this account you give is Phan Thanh Tịnh. Why you sure 2 accounts are one ??? May be this two persons have the same name. And if myth this 2 accounts are one, I hope Summer Clouds solve. Thanks 

a,b are positive number, aren't integer, so they can are positive fraction

So Lê Quốc Trần Anh is wrong

The first day of this month is not Tuesday. So may be first Tuesday is 3^rd. We let's try: 

Tuesday: 3              10              17              24              31

but 31^st is last day of a month has 31days => Unsatisfactory

So first Tuesday of this month is 2^nd

We let's try too.

Tuesday: 2               9               16               23               30

So the last Tuesday of this month is 30^th may be last day of this month. But the topic talk last day isn't Tuesday. Angain have in a year will has 7 months have 31 days.

So the last day of the month is 31 

\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{100\cdot104}\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{100}-\dfrac{1}{104}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{104}\right)=\dfrac{1}{5}\cdot\dfrac{25}{104}=\dfrac{5}{104}\)

So the value of \(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+............+\dfrac{1}{100\cdot104}\)is \(\dfrac{5}{104}\)

Worst case is Peter take out 2 red balls, 2 white balls, 2 yellow balls, all blue ball and black ball.

So Peter taken:

2 + 2 + 2 + 2 + 1 = 9 [balls]

Not three members of the same color, have to take 1 ball more

Number of balls to take are: 9 + 1 = 10 [balls]

Answer 10 balls

By theme we have:

a = 2b = 3c = 4d = 6e

and because a,b,c,d,e are the ages of 5 people, them are bigger than 0

\(\Rightarrow a⋮2,3,4,6\) and a is  smallest

=> a is least common multiple

\(LCM\left(2,3,4,6\right)=12\)

So a is 12

=> \(\left\{{}\begin{matrix}b=6\\c=4\\d=3\\e=2\end{matrix}\right.\Rightarrow a+b+c+d+e=12+6+4+3+2=27\)

So the smallest possible value of a+b+c+d+e is 27

\(A=x^2-xy+y^2>x^2-xy-xy+y^2\)

\(\Rightarrow A>x^2-2xy+y^2\)

\(\Rightarrow A>\left(x-y\right)^2\)

But \(\left(x-y\right)^2\ge0\)

\(\Leftrightarrow A\ge0\)

But \(A\ne0\Rightarrow A>0\)

B = 19 - 6x - 9x²

\(=-\left(9x^2+6x\right)+19\)

\(=-\left[\left(3x\right)^2+2\cdot3x\cdot1+1^2\right]+18\)

\(=-\left(3x+1\right)^2+18\)

\(-\left(3x+1\right)^2\le0\)

\(\Rightarrow B\le18\)

=> Maximum of B = 18 at \(-\left(3x+1\right)^2=0\) and x = \(-\dfrac{1}{3}\)

A = 2x - x²

= \(-\left(x^2-2\cdot x\cdot1+1^2\right)+1\)

\(=-\left(x-1\right)^2+1\)

\(-\left(x-1\right)^2\le0\)

\(\Rightarrow-\left(x-1\right)^2+1\le1\)

=> Maximum of A = 1 at \(-\left(x-1\right)^2=0\) and x = 1

A = x² + 8x + 1 

= x² + 2.x.4 + 16 - 15

= \(\left(x+4\right)^2-15\ge-15\)

=> Minimum of A = -15 at \(\left(x+4\right)^2=0\) and x = -4

B = x² - x + 1 

= \(x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=> Minimun of B = \(\dfrac{3}{4}\) at \(\left(x-\dfrac{1}{2}\right)^2=0\) and x = \(\dfrac{1}{2}\)

Suppose:

a > b > c > d > e

=> a - b > b - c

=> |a-b| = |b-c| when a-b = c-b or a = c

Again have:

b - c > c - d 

=> |b-c| = |c-d| when b-c = d-c or b = d

Similarly we get results:

a = c ; b = d; c = e; d = a

* a = c and d = a => a = d => a = c = d [1]

* b = d and d = a => a = b => a = b = d [2]

* c = e and a = c => a = e => a = c = e [3]

From [1], [2], [3] => a = b = c = d = e

3 x 10 + 3 x 20 

= 30 + 60

= 90

@Phan Thanh Tinh, in the subject said \(x,y\ge0\), so I only apply :]