\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\)

\(=4a^2b^2-\left(a^4+2a^2b^2+b^4\right)+\left(2b^2c^2+2c^2a^2\right)-c^4\)

\(=\left(2ab\right)^2-\left(a^2+b^2\right)^2+2c^2\left(a^2+b^2\right)-c^4\)

\(=\left(2ab\right)^2-\left[\left(a^2+b^2\right)^2-2c^2\left(a^2+b^2\right)+c^4\right]\)

\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[-\left(a^2-b^2\right)+c^2\right]\)

\(=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(a+c-b\right)\)

Because a,b,c are 3 edges of a triangle, so

a+b > c => a+b - c > 0

a+c > b => a+c - b > 0

a,b,c > 0 => a+b+c > 0

b+c > a => c-a+b > 0

So \(\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(a+c-b\right)>0\)

So \(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\)

x³ + y³ + z³ = x + y + z + 2011

=> x³ + y³ + z³ - x - y - z = 2011

=> (x³ - x) + (y³ - y) + (z³ - z) = 2011

=> x(x² - 1) + y(y² - 1) + z(z² - 1) = 2011

=> x(x-1)(x+1) + y(y-1)(y+1) + z(z-1)(z+1) = 2011

But we have:

x(x-1)(x+1); y(y-1)(y+1) and z(z-1)(z+1) are the product of three consecutive natural numbers so divisible by 3

We can prove that a simple way:

- x \(⋮3\) => x(x-1)(x+1) divisible by 3

- x divide 3 residuals 1 => x - 1divisible by 3

- x divide 3 residuals 2 => x + 1 divisible by 3

So x(x-1)(x+1) divisible by 3

So we have:

x(x-1)(x+1) + y(y-1)(y+1) + z(z-1)(z+1) \(⋮3\)

But not divisible by 3

=> There are no integer x,y,z satisfactory 

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=3^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=9\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=9\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2abc}{abc}=9\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=9\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=7\)

\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)

\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)-\left(a-c\right)^3=-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Rightarrow-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=210\)

\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)=-70\)

\(\Rightarrow\left|a-b\right|\cdot\left|b-c\right|\cdot\left|c-a\right|=70=2\cdot5\cdot7\)

But a,b,c are integer => \(\left|a-b\right|+\left|b-c\right|+\left|c-a\right|=2+5+7=14\)

\(M=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

=> The minimum of M is -36 when x = 0 or x = -5

\(1=\dfrac{1\cdot2}{2}\)

\(3=\dfrac{2\cdot3}{2}\)

\(6=\dfrac{3\cdot4}{2}\)

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

=> \(2016^{th}=\dfrac{2016\cdot2017}{2}=2033136\)

\(\overline{ab}+\overline{ba}=100\)

\(\Rightarrow10a+b+10b+a=100\)

\(\Rightarrow11a+11b=100\)

\(\Rightarrow11\left(a+b\right)=100\)

\(\Rightarrow a+b=\dfrac{100}{11}\)

But x \(\in N\)* and \(y\in N\)

=> Don't have a,b satisfactory 

ab + ba = 100

<=> 2.a.b = 100

<=> ab = 50

So ab = 50

Additional: This is a learning web, not a web devoted to curse. Ask the Admin to delete the spam question, in Vietnamese. Offer to deduct points or lock the fraudulent account that Alone mentioned above. Thank you sincerely.

4 > 3 => 4⁵⁷⁷⁷ > 3⁵⁷⁷⁷

And 3⁵⁷⁷⁷ > 3⁵⁶⁷⁴

==> 4⁵⁷⁷⁷ > 3⁵⁶⁷⁴

a.

\(x^2-2y^2=xy\)

\(\Leftrightarrow x^2-2y^2-xy=0\)

\(\Leftrightarrow\left(x^2-y^2\right)-\left(y^2+xy\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)

But \(x+y\ne0\)

\(\Rightarrow x-2y=0\Leftrightarrow x=2y\)

\(\Rightarrow P=\dfrac{x-y}{x+y}=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

I edited the subject

\(x^3-x^2-4x^2+8x-4\)

\(=x^2\left(x-1\right)-\left(4x^2-8x+4\right)\)

\(=x^2\left(x-1\right)-\left[\left(2x\right)^2-2\cdot2x\cdot2+2^2\right]\)

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

\(=x^2\left(x-1\right)-4\left(x-1\right)^2\)

\(=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\)

Oh no,

Because (−1)⋅(−2)⋅(−3)⋅...⋅(−2017) = − 2017!

=> (−1)⋅(−2)⋅(−3)⋅...⋅(−2017) < 0

\(\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot...\cdot\left(-2017\right)=-2017\text{!}\)

Tell this number is a

if a = 999 => (a-3000).3 = -6003

if a = 998 => (a-3000).3 = -6006

=> Max A is Max (a-3000).3 

So Melinda can get the largest number when he chooses number 999

=> This number is: -6003

second term is: 2² - 3 = 1

3rd term is: 1² - 3 = -2

4th term is: (-2)² - 3 = 1

....

2011th term is: 1² - 3 = -2

He will takes: 480.600 = 288000 [minutes] = 4800 hours

The answer: 180^o

\(81\cdot3+27+9\cdot2+3\cdot3+1\cdot3=300\)

So the answer is 3 + 1 + 2 + 3 + 3 = 12 [boxes]

The times of Timmy takes is:

1,5.5 = 7,5 [h]

So in an hour, Manny can mow: \(\dfrac{1}{1,5}=\dfrac{2}{3}\left(one-acre-yard\right)\)

And Timmy cam mow: \(\dfrac{1}{7,5}=\dfrac{2}{15}\left(one-acre-yard\right)\)

So they will complete them working together after:

\(\dfrac{1}{\dfrac{2}{3}+\dfrac{2}{15}}=\dfrac{5}{4}\left(hours\right)\)

\(\dfrac{5}{4}h=75m\)

Answer: 75 minutes