The question must be changed to: "How much does the small boards and big boards cost all".

They all bought: \(5+3+1+2+4+6=21\left(big\right)\) boards.

They all bought: \(2+4+6+5+3+1=21\left(small\right)\) boards, too.

So it all cost: \(21.\left(486000+88000\right)=12054000\left(dollars\right)\).

So the answer is: $12.054.000

Using Pythagore's Theorem we have : 

\(24^2+h^2=72^2\rightarrow h^2=4608\Rightarrow h=\sqrt{4608}\)

Using calculator we get a height of about \(67,88\left(meter\right)\)

\(5^x+5^{x+1}=750\)

\(\Rightarrow5^x\left(1+5^1\right)=750\)

\(\Rightarrow5^x\cdot6=750\)

\(\Rightarrow5^x=\dfrac{750}{6}=125\)

\(\Rightarrow x=3\)

There are 56 triangles in the figure

There are: \(3\cdot4\cdot5=60\left(combinations\right)\) that she can watch.

there are 3*4*5=60

it is correct?

I'm sorry.Let a,b>0....

We have:

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

Other way:

\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

From (*) and (**) we have  \(a+b\le2\)

Your ex is so hard to do :) 

We have:

a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)

Other way:

a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2

=(a+12b)2+34b2=(a+12b)2+34b2

Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b

34b2≥034b2≥0 with ∀b∀b

So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b

or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)

From (*) and (**) we have  a+b≤2a+b≤2

Your ex is so hard to do :) 

From what 100% Vietnamese people are misread?

Answer: It's "wrong word"