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Wie löst man diese Aufgabe? Von 3 Zahlen a,b,c weiss man: ggt(a,b)=10, ggt (b,c)=15, kgv (a,b)=2160, und kgv (a,c)=1200. Berechne a,b,c.

$\small{ \boxed{~ \begin{array}{rcl} ggt (a,b)=10 &=& 2^1\cdot 3^0 \cdot 5^1 \\ \qquad \text{ die mimimalsten Exponenten von a und b werden überommen}\\ ggt (b,c)=15 &=& 2^0\cdot 3^1 \cdot 5^1 \\ \qquad \text{ die mimimalsten Exponenten von b und c werden überommen}\\ kgv (a,b)=2160 &=& 2^4\cdot 3^3 \cdot 5^1 \\ \qquad \text{ die maximalsten Exponenten von a und b werden überommen}\\ kgv (a,c)=1200 &=& 2^4\cdot 3^1 \cdot 5^2 \\ \qquad \text{ die maximalsten Exponenten von a und c werden überommen}\\ a\cdot b = ggt (a,b)\cdot kgv (a,b)= 10\cdot 2160 = 21600 &=& 2^5\cdot 3^3 \cdot 5^2 \\ \end{array} ~} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{?}\cdot 3^{?}\cdot 5^{?} \\ b &=& 2^{?}\cdot 3^{?}\cdot 5^{?} \\ c &=& 2^{?}\cdot 3^{?}\cdot 5^{?} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{ggt (a,b)=10 = 2^1\cdot \ \cdots } \text{ ist gerade. }\\ \mathbf{kgv (a,b)=2160 = 2^4\cdot \ \cdots } \text{ ist gerade: }\\ \text{Der mimimalste Exponent von a und b ist } 2^1. \\ \text{Der maximalste Exponent von a und b ist } 2^4. \\ \rightarrow a \text{ und } b \text{ müssen gerade sein mit einem Primfaktor von } 2^{4\text{ oder } 1} \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4\text{ oder } 1}\cdot 3^{?}\cdot 5^{?} \\ b &=& 2^{1\text{ oder } 4}\cdot 3^{?}\cdot 5^{?} \\ c &=& 2^{?}\cdot 3^{?}\cdot 5^{?} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{ggt (b,c)=15 = 2^0\cdot 3^1 \cdot 5^1 } \text{ ist ungerade. }\\ \text{Da b gerade ist, muss c ungerade sein. c hat nur ungerade Primfaktoren. }\\ \text{Der Primfaktor 2 verschwindet mit }2^0 = 1. \quad 2^0 \text{ stammt von } c \\ \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4\text{ oder } 1}\cdot 3^{?}\cdot 5^{?} \\ b &=& 2^{1\text{ oder } 4}\cdot 3^{?}\cdot 5^{?} \\ c &=& 2^{0}\cdot 3^{?}\cdot 5^{?} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{kgv (a,c)=1200 = 2^4\cdot 3^1 \cdot 5^2 } :\\ \text{Der maximale Exponent von a und c }~ 2^4 \text{ muss von a stammen, }\\ \text{da c ungerade ist und den Exponenten } 2^0 \text{ hat. }\\ \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4}\cdot 3^{?}\cdot 5^{?} \\ b &=& 2^{1}\cdot 3^{?}\cdot 5^{?} \\ c &=& 2^{0}\cdot 3^{?}\cdot 5^{?} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{ggt (a,b)=10 = \cdots 5^1 } \\ \mathbf{kgv (a,b)=2160 = \cdots 5^1 }: \\ \text{Der mimimalste Exponent von a und b ist } 5^1. \\ \text{Der maximalste Exponent von a und b ist } 5^1. \\ \mathbf{a\cdot b = 21600 = 2^5\cdot 3^3 \cdot 5^2 } :\\ 5^1 \cdot 5^1 = 5^2 \\ \qquad 5^0 \cdot 5^2 = 5^2 \text{ und } 5^0 \cdot 5^2 = 5^2 \text{ kommen nicht in Frage,}\\ \text{da der mimimalste Exponent von a und b }~ 5^1 \text{ ist.}\\ \text{Sowohl a alsauch b müssen einen Primfaktor von } 5^1 \text{ haben. } \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4}\cdot 3^{?}\cdot 5^{1} \\ b &=& 2^{1}\cdot 3^{?}\cdot 5^{1} \\ c &=& 2^{0}\cdot 3^{?}\cdot 5^{?} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{kgv (a,c)=1200 = 2^4\cdot 3^1 \cdot 5^2 } :\\ \text{Der maximale Exponent von a und c }~ 5^2 \text{ muss von c stammen, }\\ \text{da a den Exponenten } 5^1 \text{ hat. }\\ \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4}\cdot 3^{?}\cdot 5^{1} \\ b &=& 2^{1}\cdot 3^{?}\cdot 5^{1} \\ c &=& 2^{0}\cdot 3^{?}\cdot 5^{2} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{ggt (a,b)=10 = \cdots \ \cdot 3^0\cdot \ \cdots } \\ \mathbf{kgv (a,b)=2160 = \cdots \ \cdot 3^3\cdot \ \cdots }:\\ \text{Der mimimalste Exponent von a und b ist } 3^0. \\ \text{Der maximalste Exponent von a und b ist } 3^3. \\ \rightarrow a \text{ und } b \text{ müssen einen Primfaktor von } 3^{0\text{ oder } 3} \text{ haben. } \end{array} }$

$\small{ \color{red} \boxed{~ \begin{array}{rcl} a &=& 2^{4}\cdot 3^{0\text{ oder } 3}\cdot 5^{1} \\ b &=& 2^{1}\cdot 3^{3\text{ oder } 0}\cdot 5^{1} \\ c &=& 2^{0}\cdot 3^{?}\cdot 5^{2} \\ \end{array} ~} }$

$\small{ \begin{array}{lrcl} \mathbf{kgv (a,c)=1200 = \cdots \ \cdot 3^1\cdot \ \cdots }: \\ \text{Da b nicht den Primfaktor } 3^1 \text{ haben kann,}\\ \text{muss der Primfaktor } 3^1 \text{ von c stammen }\\ \text{Der maximalste Exponent von a und c ist }~ 3^1 \\ \text{Also kann a nur den Primfaktor }~ 3^0 \text{ und nicht }~3^3\\ \text{haben, den muss dann b haben. } \end{array} }$

$\small{ \boxed{~ \begin{array}{rclcr} a &=& 2^{4}\cdot 3^{0}\cdot 5^{1} = 16 \cdot 5 &=& 80 \\ b &=& 2^{1}\cdot 3^{3}\cdot 5^{1} = 2 \cdot 27 \cdot 5 &=& 270 \\ c &=& 2^{0}\cdot 3^{1}\cdot 5^{2} = 3 \cdot 25 &=& 75\\ \end{array} ~} }$

0,5*e^(X)-2 =2-e^X

Ich vermute die Gleichung heißt: $0,5\cdot e^x-2 = 2-e^x$

$\small{ \begin{array}{rcll} 0,5\cdot e^x-2 &=& 2-e^x \qquad & | \qquad +2 \\ 0,5\cdot e^x &=& 4 \qquad & | \qquad +e^x \\ 0,5\cdot e^x +e^x &=& 4\\ 1,5\cdot e^x &=& 4 \qquad & | \qquad :1,5 \\ e^x &=& \frac{4}{1,5} \\ e^x &=& \frac{4\cdot 2}{3} \\ e^x &=& \frac{8}{3} \qquad & | \qquad \ln{()} \\ \ln{(e^x)} &=& \ln{( \frac{8}{3})} \\ x\cdot \ln{(e)} &=& \ln{( \frac{8}{3})} \qquad & | \qquad \ln{(e)} = 1 \\ x &=& \ln{( \frac{8}{3})} \\ x &=& \ln{(8)} - \ln{(3)} \\ x &=& 2,07944154168 - 1,09861228867 \\ x &=& 0,98082925301 \\ \end{array} }$

x^5=7

$x^5=7 \\ x = \sqrt[5]{7}\\ x = 1.47577316159$

eliminate "vi" from the following equations 1)vf - vi + at S=vit+1/2at^2

$\begin{array}{lrcl} (1) & v_f &=& v_i + a\cdot t \\ & v_i &=& v_f - a\cdot t \\\\ (2) & s &=& v_i \cdot t +\frac12 a\cdot t^2 \\ & s &=& (v_f - a\cdot t ) \cdot t +\frac12 a\cdot t^2 \\ & s &=& v_f\cdot t - a\cdot t \cdot t +\frac12 a\cdot t^2 \\ & s &=& v_f\cdot t - a\cdot t^2 +\frac12 a\cdot t^2 \\ & \mathbf{s} & \mathbf{=} & \mathbf{v_f\cdot t - \frac12 a\cdot t^2} \\ \end{array}$

If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

3n + 1 is a perfect square:

$\small{ \begin{array}{rcl} 3n+1 &=& a^2 \\ 3n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{3}\\ \hline n+1 &=& \frac{a^2-1}{3} +1 \\ n+1 &=& \frac{a^2-1+3}{3}\\ n+1 &=& \frac{a^2+2}{3}\\ \end{array} }$

Because $\frac{a^2+2}{3}$ is a integer then $a^2+2$ is divisible by 3, then $a^2$ is not divisible by 3  and also $a$ is not divisible by 3,

because if 3 is not a prime factor in $a^2$ a perfect spuare, then 3 is not a prime factor in  $a$

Two numbers are not divisible by 3. It is  $3b +1$ and $3b + 2$

1. We substitute $a = 3b+1$

$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+1\\ n+1 &=& \frac{(3b+1)^2+2}{3} \\ n+1 &=& \frac{9b^2+6b+1+2}{3} \\ n+1 &=& \frac{9b^2+6b+3}{3} \\ n+1 &=& 3b^2+2b+1 \\ n+1 &=& b^2 + b^2 + b^2 +2b+1 \\ n+1 &=& b^2 + b^2 + (b+1)^2\\ \end{array} }$

So n + 1 is the sum of three perfect squares and $b = \frac{a-1}{3}$.

2. We substitute $a = 3b+2$

$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+2\\ n+1 &=& \frac{(3b+2)^2+2}{3} \\ n+1 &=& \frac{9b^2+12b+4+2}{3} \\ n+1 &=& \frac{9b^2+12b+6}{3} \\ n+1 &=& 3b^2+4b+2 \\ n+1 &=& b^2 + b^2 + b^2 +2b+2b+1+1 \\ n+1 &=& b^2 + b^2+2b+1 + b^2 +2b+1 \\ n+1 &=& b^2 +(b+1)^2+(b+1)^2\\ \end{array} }$

So n + 1 is the sum of three perfect squares and $b = \frac{a-2}{3}$.

Example 1:

$\small{ \begin{array}{rcl} a &=&7 \\ 3n+1 =a^2&=& 7^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{49-1}{3} = 16\\\\ n+1 &=& 16+1=17 \\ 17 &=& b^2+b^2+(b+1)^2 \qquad a=3b+1 \qquad \rightarrow b = \frac{a-1}{3} = \frac{7-1}{3} = 2\\ 17 &=& 2^2+2^2+3^2 = 4+4+9\\ \end{array} }$

Example 2:

$\small{ \begin{array}{rcl} a &=&8 \\ 3n+1 =a^2&=& 8^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{64-1}{3} = 21\\\\ n+1 &=& 21+1=22 \\ 22 &=& b^2+(b+1)^2+(b+1)^2 \qquad a=3b+2 \qquad \rightarrow b = \frac{a-2}{3} = \frac{8-2}{3} = 2\\ 22 &=& 2^2+3^2+3^2 = 4+9+9\\ \end{array} }$

Hello,

I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!

Let f(x)= ax / (2x+3)

Investigate if you can define the 'a' so that f(f(x))=x

New edit, without mistake:

$\begin{array}{rcl} a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ -3 } \end{array}$

Hello,

I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!

Let f(x)= ax / (2x+3)

Investigate if you can define the 'a' so that f(f(x))=x

$\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}$

If n is a positive integer such that 2n + 1 is a perfect square, show that n + 1 is the sum of two successive perfect squares.

2n + 1 is a perfect square:

$\begin{array}{rcl} 2n+1 &=& a^2 \\ 2n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{2}\\ \hline n+1 &=& \frac{a^2-1}{2} +1 \\ n+1 &=& \frac{a^2-1+2}{2}\\ n+1 &=& \frac{a^2+1}{2}\\ \end{array}$

Because $\frac{a^2+1}{2}$ is a integer then $a^2+1$ is even and divisible by 2, then $a^2$ is odd or also $a$ is odd.

A odd number is  $2b+1$

$\begin{array}{rcl} n+1 &=& \frac{a^2+1}{2}\qquad \text{substitute }\ a = 2b+1\\ n+1 &=& \frac{(2b+1)^2+1}{2} \\ n+1 &=& \frac{4b^2+4b+1+1}{2} \\ n+1 &=& \frac{4b^2+4b+2}{2} \\ n+1 &=& 2b^2+2b+1 \\ n+1 &=& b^2 + b^2+2b+1 \\ n+1 &=& b^2 +(b+1)^2\\ \end{array}$

So n + 1 is the sum of two successive perfect squares and $b = \frac{a-1}{2}$.

Example:

$\begin{array}{rcl} a &=& 5\\ 2n+1 =a^2 &=& 25 \qquad \rightarrow\qquad n = \frac{a^2-1}{2} = \frac{25-1}{2} = 12\\\\ b &= &\frac{a-1}{2} = \frac{5-1}{2} = 2\\ n+1 &=& 12+1 = 13 \\ 13 &=& b^2 + (b+1)^2 = 2^2+3^2 = 4+9\\ \end{array}$

Find the 2015th term of the geometric sequence 1,−1/3,-1/6,−1/9 ,.

In a geometric sequence, the common ratio, r between any two consecutive terms is always the same.

$\begin{array}{rcl} \dfrac{ \text{Any term} } { \text{Previous term}} &=& \dfrac{u_n}{ u_n-1} = \text{ constant } = r \\ \end{array}$

$\begin{array}{rcr} 1,-\frac13,-\frac16,-\frac19,-\cdots \\ u_1 &=& 1\\ u_2 &=& -\frac13\\ u_3 &=& -\frac16\\ u_4 &=& -\frac19\\ \cdots \\ \end{array}$

If three terms,  $u_n,\ u_{n+1},\ u_{n+2}$ are in geometric sequence, then:

$\begin{array}{rcl} \dfrac{ u_{n+2} } { u_{n+1} } &=& \dfrac{ u_{n+1} } { u_{n} }\\ \end{array}$

We have

$\begin{array}{rcr} \dfrac{ u_2 } { u_1 } = \dfrac{ -\frac13 } { 1 } = -\frac13 &\ne& \dfrac{ u_3 } { u_2 } = \dfrac{ -\frac16 } { -\frac13 } = \frac12 \\ &\ne& \dfrac{ u_4 } { u_3 } = \dfrac{ -\frac19 } { -\frac16 } = \frac23 \\ \end{array}$

Sorry, this is not a geometric sequence.

wieviel prozent sind 3 stück von 10000 stück ?

$\begin{array}{rcl} \dfrac{3}{10000} &=& 0,0003 \\\\ \dfrac{3}{10000} &=& \dfrac{ 0,03 } {100} \\\\ \dfrac{3}{10000} &=& 0,03\cdot \dfrac{ 1 } {100} \\\\ \dfrac{3}{10000} &=& 0,03\cdot \% \\\\ 3 &=& 10000 \cdot 0,03\ \% \\\\ 3\ \text{ Stück sind }\ 0,03\ \% \text{ von }\ 10000\\ \end{array}$