P = 43 . 36 = (22)3.36=26.36=(2.3)6=66

P = 4³ . 3⁶ = \(\left(2^2\right)^3.3^6=2^6.3^6=\left(2.3\right)^6=6^6\)

P = 43. 36 = (22) 3,36 = 26,36 = (2,3) 6 = 66

The factorials that are divisible by 9 is 9!, 10!, 11!, 12!,..., 100! (These factorials contains factor 9)

=> There are: (100 - 9) + 1 = 92 (factorials)   

Lê Anh Duy is wrong.

\(6!=1.2.3.4.5.6=9.1.2.4.5.2⋮9\)

So there are: 92+1=93(factorials)

Land : A = 1.3 + 3.5 + 5.7 + ...... + 99.101

=> 4A = 1.3.(5 - 1) + 3.5.(7 - 3) + ...... + 99.101.(103 - 99)

=> 4A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ...... + 99.100.101

=> 4A = 99.100.101

=> A = 99.100.101/4

=> A = 249975

Land : A = 1.3 + 3.5 + 5.7 + ...... + 99.101

=> 4A = 1.3.(5 - 1) + 3.5.(7 - 3) + ...... + 99.101.(103 - 99)

=> 4A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ...... + 99.100.101

=> 4A = 99.100.101

=> A = 99.100.101/4

=> A = 249975

Put A = 1.3 + 3.5 + 5.7 + ...... + 99.101

=> 4A = 1.3.(5 - 1) + 3.5.(7 - 3) + ...... + 99.101.(103 - 99)

=> 4A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ...... + 99.100.101

=> 4A = 99.100.101

=> A = 99.100.101/4

=> A = 249975

We have: \(\dfrac{x^2\left(x-3\right)}{x-9}< 0\)

\(\Rightarrow x^2\left(x-3\right)< 0\)

But \(x^2\ge0\) so \(x-3< 0\)

\(\Rightarrow x< 3\)

So \(x< 3\)

First, \(x^2\left(x-3\right)\ne0\) and \(x-9\ne0\Leftrightarrow x\ne0,3,9\)

Then we have \(\left\{{}\begin{matrix}x^2>0\\\dfrac{x^2\left(x-3\right)}{x-9}< 0\end{matrix}\right.\Leftrightarrow\dfrac{x-3}{x-9}< 0\)

We also have \(x-9< x-3\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x-9< 0\end{matrix}\right.\)

\(\Leftrightarrow3< x< 9\)

I think we should add condition:

That is: x - 9 \(\ne\)0 <=> x \(\ne\)9

By applying the property of sequence of equivalent ratios,we have :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{12}{5}=2.4\)\(\Rightarrow\left\{{}\begin{matrix}x=2.4\times2=4.8\\y=2.4\times3=7.2\end{matrix}\right.\)

I play soccer and play games

Very good

a)\(x^2+3x+7\)

\(=x^2+3x+\dfrac{9}{4}+\dfrac{19}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

When \(x=-\dfrac{3}{2}\)

b)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(x^2-6xy+4x+9y^2-12y+4\right)+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

When \(x=5;y=\dfrac{7}{3}\)