heureka

What is x^3 + 1/8 when completely factored

$$x^3+\frac{1}{8}\\ \\
=x^3+\frac{1}{2^3}\\ \\
=x^3+ ( \frac{1}{2} ) ^3 \quad | \quad \textcolor[rgb]{1,0,0}{a^3+b^3 = (a+b)(a^2-ab+b^2) } \\\\
a = x \text{ and } b = \frac{1}{2} \\\\
x^3+ ( \frac{1}{2} ) ^3 = (x+\frac{1}{2})(x^2-x*\frac{1}{2}+(\frac{1}{2} )^2 )$$

30! [mod 899] ?

$$30! \mod 899 \stackrel{?}{=}$$

$$\small{\text{
$ 899 = 29 * 31 = p*q \quad | \quad $ let $ p = 29 $ and $ q=31 $ so \textcolor[rgb]{1,0,0}{p and q are relatively prim!}
}}$$

I.  $$\small{\text{
$ \textcolor[rgb]{0,0,1}{30! \mod p = 0} , $ because $ p = 29 $ is divider of $ 30! \ (30! = 30*\textcolor[rgb]{1,0,0}{29}*28*...3*2*1)
$
}}$$ 

$$\small{\text{
$ \textcolor[rgb]{0,0,1}{30! \mod p = r} , $ if $ p = 29 $ we have $ r = 0
$
}}$$

II. $$\small{\text{
$ \textcolor[rgb]{0,0,1}{30! \mod q = s } \quad q=31 $ is a prime number so $ \textcolor[rgb]{0,0,1}{(31-1)! \equiv -1 \mod 31} $ [Wilson]
}}$$

$$\small{\text{
$ (31-1)! \equiv -1 + 31 \mod 31 $ or $ 30! \equiv \textcolor[rgb]{0,0,1}{30} \mod 31 $ we have $ s=30 $
}}$$

III. 

Since p  and q  are relatively prime, there are integers a  and b  such that  ap+bq=1. You can find a  and b  using the Extended Euclidean algorithm.

what are the last 2 digits of 7^2800  or what is $$7^{2800} \mod 100$$ ?

$$7^{2800} \mod 100 \quad | \quad 2800 = 2^4*175\\
= 7^{ 16 * 175 } \mod 100 \\
= ( 7^{16} )^{175} \mod 100 \quad | \quad 7^{16}= (7^4)^4\\
= ( (7^4)^4 )^{175} \mod 100 \quad | \quad 7^{4}= (7^2)^2\\
= ( ( (7^2)^2 )^4 )^{175} \mod 100 \quad | \quad 7^2 = 49\\
= ( ( (49)^2 )^4 )^{175} \mod 100 \quad | \quad 49^2 = 2401\\
= ( ( 2401 )^4 )^{175} \mod 100 \quad | \quad 2401 \mod 100 = 1 \quad\textcolor[rgb]{1,0,0}{ x^{a*b} \mod d = (x^a\mod d)^b \mod d }\\
= ( ( 1 )^4 )^{175} \mod 100 \quad | \quad 1^4 = 1 \\
= ( 1 )^{175} \mod 100 \quad | \quad 1^{175} = 1 \\
= 1 \mod 100 \\\\
7^{2800} \mod 100 = 1 \mod 100 \quad | \quad 1 \mod 100 = 01$$

the last 2 digits of 7^2800 = 01

Simplify the radical expression :

1.)  (-1-√2)/(1+√2)

$$\frac{(-1- \sqrt{2} )}{ (1+ \sqrt{2} ) }
=-\frac{(1+ \sqrt{2} )}{ (1+ \sqrt{2} ) } = -1$$

2.)  (√8)/(√2+3)

$$\frac { \sqrt{8} } { \sqrt{2} + 3} = \frac { 2\sqrt{2} } { \sqrt{2} + 3} *\frac{ \sqrt{2}-3 }{ \sqrt{2}-3 } =-\frac{2}{7}(2-3\sqrt{2})$$

3.)  (√3)/(√27-√3)

$$\frac{ \sqrt{3} }{ \sqrt{27}-\sqrt{3} } = \frac{ \sqrt{3} }{ 3\sqrt{3}-\sqrt{3} } = \frac{ \sqrt{3} }{ 2\sqrt{3} } = \frac{1}{2}$$

4.)  (√2-√8)/2(√2-1)

$$\frac{ \sqrt{2}-\sqrt{8} }{ 2(\sqrt{2}-1) } =\frac{ \sqrt{2}-2\sqrt{2} }{ 2(\sqrt{2}-1) } =\frac{ -\sqrt{2} }{ 2( \sqrt{2}-1) } * \frac{ \sqrt{2}+1 }{ \sqrt{2}+1 } = \frac{ -\sqrt{2} } {2} ( \sqrt{2} + 1) = - ( 1 + \frac{ \sqrt{2} } {2} )$$

5.)  (3√3-2√2)/(√3-√2)

$$\frac{ 3\sqrt{3}-2\sqrt{2} } { \sqrt{3}-\sqrt{2} } * \frac{\sqrt{3}+\sqrt{2}}{ \sqrt{3}+\sqrt{2} } = ( 3\sqrt{3}-2\sqrt{2} ) * ( \sqrt{3}+\sqrt{2} ) = 9 + \sqrt{6} - 4 = 5 + \sqrt{6}$$

6.)  (√2-√8)/(2√2-1)

$$\samll{\text{
$
\frac{ \sqrt{2}-\sqrt{8} }{ 2\sqrt{2}-1 }
=\frac{ \sqrt{2}-2\sqrt{2} }{ 2\sqrt{2}-1 }
=\frac{ -\sqrt{2} }{ 2\sqrt{2}-1 } * \frac{ 2\sqrt{2}+1 }{ 2\sqrt{2}+1 }
= \frac{ -\sqrt{2} } {7} ( 2\sqrt{2} + 1)
= -\frac{ 1 } {7} ( 4 + \sqrt{2} )
$
}}$$

7.)  (1-√2)/(1+√2)

$$\small{\text{
$
\frac{ 1-\sqrt{2} } { 1+\sqrt{2} } *\frac{ 1-\sqrt{2} }{ 1-\sqrt{2} }
= -( 1-\sqrt{2} )( 1-\sqrt{2} ) = -(1-2\sqrt{2}+2) = -(3-2\sqrt{2}) = 2\sqrt{2}-3
$
}}$$

again 5.) (3√3-2√2)/(√3-√2)

Quelle est la limite de la suite U_{n = √n - √(n-1)}

10^100 is called  a Googol

wie berechne ich die Nullstellen bei 4x² - 12x = 0

4x kann vor die Klammer gesetzt werden : 4x(x-3) = 0

Da die Gleichung gleich der Null gesetzt wird, können wir sagen:

Beide Faktoren   4x   oder   x-3  ergeben den Wert Null für die Gleichung, wenn sie  auf Null gesetzt werden.

0 * ( x-3 ) = 0       oder      4x * ( 0 ) = 0

So die 1. Lösung ergibt sich, wenn 4x = 0 gesetzt wird. x wäre dann 0 (x=0)

Die 2. Lösung ergibt sich, wenn x-3 = 0 gesetzt wird. x wäre dann 3 (x=3)

Die beiden Nullstellen lauten:  $$x_1 = 0$$ und $$x_2 = 3$$

Probe: 4 * 0 * 0 - 12 * 0 = 0   richtig!

          4 * 3 * 3 - 12 * 3 = 36 - 36 = 0   richtig!

$${\frac{\left({{\mathtt{10}}}^{{\mathtt{102}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{10}}}^{{\mathtt{100}}}\right)}{{{\mathtt{10}}}^{{\mathtt{100}}}}}$$=?

$$\dfrac{(10^{102}+10^{100}) }{ 10^{100} } \\\\
= \dfrac{(10^{100}* 10^{2}+ 10^{100} ) }{ 10^{100} }\\\\
= \dfrac{ 10^{100}* (10^{2}+ 1 ) }{ 10^{100} }\\\\
= \dfrac{ 10^{100} }{ 10^{100} } *(10^{2}+ 1 ) \\\\
=10^{2}+ 1\\
=100+1 \\
=101$$

$${\frac{\left({\frac{{\mathtt{1}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{{\mathtt{y}}}^{{\mathtt{2}}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{y}}}}\right)}}$$ = ?

$$\dfrac{
\left( \dfrac{1}{x^2} + \dfrac{1}{y^2} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)\left( \dfrac{1}{x} + \dfrac{1}{y} \right) -2\times\dfrac{1}{x}\times\dfrac{1}{y}
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}\\ \\ \\
=
\dfrac{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}-
\dfrac{
\dfrac{2}{xy}}
{
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{xy\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
}
\\ \\ \\
=
\left( \dfrac{1}{x} + \dfrac{1}{y} \right)
-
\dfrac{2}{x+y}$$

(2x-10)(x+5)=x^2+10x-75

okay:  x^2-10x=-25

now: $$x^2-10x+25=0$$

use the abc - formula:

if $$ax^2+bx+c = 0$$ than $$x_{1,2} = \frac{1}{2a}*(b-\sqrt{b^2-4*a*c})$$

$$1x^2 -10x+25=0 \\
\underbrace{1}_{a=1}x^2 \underbrace{-10}_{ b=-10}x+\underbrace{25}_{c=25}=0 \\\\
x_{1,2} = \frac{1}{2*1}*(10-\sqrt{100-4*1*25})\\\\
x_{1,2} = \frac{1}{2}*(10-\sqrt{100-100})\\\\
x = \frac{1}{2}*(10-0)\\\\
x = \frac{1}{2}*10\\\\
x = \frac{10}{2}\\\\
x = 5$$