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$A = x^{2} - x y + y^{2} > x^{2} - x y - x y + y^{2}$

$\Rightarrow A > x^{2} - 2 x y + y^{2}$

$\Rightarrow A > {(x - y)}^{2}$

But ${(x - y)}^{2} \geq 0$

$\Leftrightarrow A \geq 0$

But $A \neq 0 \Rightarrow A > 0$

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$A = x^{2} - x y + y^{2} > x^{2} - x y - x y + y^{2}$

$\Rightarrow A > x^{2} - 2 x y + y^{2}$

$\Rightarrow A > {(x - y)}^{2}$

But ${(x - y)}^{2} \geq 0$

$\Leftrightarrow A \geq 0$

But $A \neq 0 \Rightarrow A > 0$

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$x^{2} + 2 x y + 6 x + 6 y + 2 y^{2} + 8 = 0$

$\Leftrightarrow x^{2} + y^{2} + 9 + 2 x y + 6 y + 6 x + y^{2} - 1 = 0$

$\Leftrightarrow {(x + y + 3)}^{2} + y^{2} = 1$

$y^{2} \geq 0 \Rightarrow {(x + y + 3)}^{2} \leq 1 \Rightarrow - 1 \leq x + y + 3 \leq 1$

$\Leftrightarrow 2012 \leq x + y + 2016 \leq 2014$

Hence :

B = 2012 only when : y = 0 ; x = 2012 - 2016 = -4

B = 2014 only when : y = 0 ; x = 2014 - 2016 = -2

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We have: 5x = ...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

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a) a2 + 2(a + 1)2 + 3(a + 2)2 + 4(a + 3)2

= a2 + 2(a2 + 2a + 1) + 3(a2 + 4a + 4) + 4(a2 + 6a + 9)

= a2 + 2a2 + 4a + 2 + 3a2 + 12a + 12 + 4a2 + 24a + 36

= 10a2 + 40a + 50 = a2 + 10a + 25 + 9a2 + 30a + 25

= (a + 5)2 + (3a + 5)2

b) (a + b)(a - b) + (b + c)(b - c) = a2 - b2 + b2 - c2

= a2 - c2 = (a + c)(a - c)

c) (ax + by)2 + (ay - bx)2

= a2x2 + 2axby + b2y2 + a2y2 - 2axby + b2x2

= a2x2 + a2y2 + b2x2 + b2y2

= a2(x2 + y2) + b2(x2 + y2) = (a2 + b2)(x2 + y2)

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122 x 71 = (112 + 1)(11 x 6 + 5)

= 113 x 6 + 112 x 5 + 11 x 6 + 5 $\equiv 5 (m o d 11)$

So, the answer is 5

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Let x be the least satisfied integer (x > 6).

x leaves a remainder of 6 when divided by 7 and 11, so x - 6 is divisible by 7 and 11 or $x - 6 ⋮ 77$

$\Rightarrow x - 6 = 77 \Rightarrow x = 83$

So, the answer is 83

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a) $x^{4} + 4 = x^{4} + 4 x^{2} + 4 - 4 x^{2} = {(x^{2} + 2)}^{2} - {(2 x)}^{2}$

$= (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$

b) $4 x^{4} + 81 = 4 x^{4} + 36 x^{2} + 81 - 36 x^{2} = {(2 x^{2} + 9)}^{2} - {(6 x)}^{2}$

$= (2 x^{2} - 6 x + 9) (2 x^{2} + 6 x + 9)$

c) $4 x^{4} - 9 x^{2} + 81 = 4 x^{4} + 36 x^{2} + 81 - 45 x^{2}$

$= {(2 x^{2} + 9)}^{2} - {(\sqrt{45} x)}^{2} = (2 x^{2} - \sqrt{45} x + 9) (2 x^{2} + \sqrt{45} x + 9)$

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Let n be the of edges of the polygon and (n ∈ N, n > 2)

We have :

180 . (n - 2) = 1260 <=> n - 2 = 7 => n = 9

Answer: 9 sides

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The number of donuts Jon has is : 29 x 51 = 1479 (donuts)

Since 1479 = 123 x 12 + 3, three donuts will be left over when he groups them again

So, the answer is 3

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The number of 1-digit numbers is : (9 - 1) + 1 = 9

The number of 2-digit numbers is : (99 - 10) + 1 = 90

The number of 3-digit numbers is : (321 - 100) + 1 = 222

The number of digits she wrote is :

9 x 1 + 90 x 2 + 222 x 3 = 855

Answer : 855

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Answer

20+30+40+50+60+70

=(20+70)+(30+60)+(40+50)

=90 + 90 +90

=270

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Answer

a, 1 + 2 + 3 + 4 5 + 6 + 7 + 8 + 9

= (1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5

= 10 + 10 + 10 + 10 + 5

= 40 + 5

=45

b, 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) x 10

= 45 x 10

= 450

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S(48) = 48 + 24 + 16 + 12 + 8 + 6 + 4 + 3 + 2 + 1 = 124

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Answer of Dao Trong Luan must be $\frac{3.4 : 2}{2}$ to $\frac{3.4 : 2}{3}$ .

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he number of 1-digit numbers is : (9 - 1) + 1 = 9

The number of 2-digit numbers is : (99 - 10) + 1 = 90

The number of 3-digit numbers is : (321 - 100) + 1 = 222

The number of digits she wrote is :

9 x 1 + 90 x 2 + 222 x 3 = 855

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Let x be the least number of members in the band, then x is divided by 3, 5, 7, 9 with remainders 1, 2, 3, 4 respecitvely.

Since 3 = 1.2 + 1 ; 5 = 2.2 + 1 ; 7 = 3.2 + 1 ; 9 = 4.2 + 1,

2x + 1 is divisible by 3 ; 5 ; 7 ; 9.

So, 2x + 1 = LCM(3,5,7,9) = 315 or x = 157

Hence, the answer is 157

We need to

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Let p, n, d, q be the number of pennies, nickels, dimes, quarters we need to make 67 cents respectively ( $p, n, d, q \in Z^{+}$ ), then p + 5n + 10d + 25q = 67

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The general forms of the satisfied palindromes are $\bar{b b}, \bar{c d c}, \bar{e f f e}$

The number of 2-digit palindromes is : (99 - 11) : 11 + 1 = 9

There are 9 choices to choose c and 10 choices to choose d, so the number of 3-digit palindromes is : 9 x 10 = 90

The only 4-digit palindrome is 1001

So, the answer is : 9 + 90 + 1 = 100

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Calling the two-digit number written by Anna was: ab then the number Ben wrote was abab
The structural analysis of the number we are:
Abab = 100 x ab + ab = 101 x ab
The result of the division is: abab: ab = 101 x ab: ab = 101