heureka

Wie geht dass, ich weiss nicht wie man da vor geht 3(9-y)=5(y-9)

1. Möglichkeit: Beide Seiten ausklammern und nach y auflösen:

\(\small{ \begin{array}{rcl} 3(9-y)&=& 5(y-9) \\ 3\cdot 9 - 3\cdot y &=& 5\cdot y - 5\cdot 9 \\ 27 - 3y &=& 5y - 45 \qquad | \qquad +3y\\ 27 &=& 8y -45 \qquad | \qquad +45\\ 45+27 &=& 8y \\ 72 &=& 8y \\ 8y &=& 72 \qquad | \qquad :8\\ y &=& \frac{72}{8} \\ \mathbf{y} & \mathbf{=} & \mathbf{9 } \end{array} }\)

2. Möglichkeit:

\(\small{ \begin{array}{rcl} 3(9-y)&=& 5(y-9) \\ 3(9-y)&=& -5(9-y) \qquad | \qquad +5(9-y)\\ 3(9-y) + 5(9-y) &=& 0 \\ 8(9-y) &=& 0 \qquad | \qquad :8\\ 9-y &=& 0 \qquad | \qquad +y\\ \mathbf{y} & \mathbf{=} & \mathbf{9 } \end{array} } \)

Kann mir bitte einer erklären wie ich mit hilfe der binomischen formeln diesen ausdruck vereinfachen kann :

(64a²- 32ab +4b²) / (4a-b)  ?

\(\small{ \begin{array}{rcl} \frac{64a^2- 32ab +4b^2}{ 4a-b } \\ &=& \frac{ 4(16a^2- 8ab +b^2)}{ 4a-b } \\ &=& 4 \cdot \left( \frac{ (4a)^2- 2\cdot 4\cdot ab +b^2}{ 4a-b } \right)\qquad | \qquad (4a)^2- 2\cdot 4\cdot ab +b^2 = (4a-b)^2\\ &=& 4 \cdot \left( \frac{ (4a-b)^2 }{ 4a-b } \right)\\ &=& 4 \cdot \left( \frac{ (4a-b)(4a-b) }{ 4a-b } \right)\\ \mathbf{ \frac{64a^2- 32ab +4b^2}{ 4a-b } } & \mathbf{=} & \mathbf{ 4 \cdot (4a-b ) } \\ \end{array} }\)

Nullstellen von f(x)= 0,5x² -3  ?

\(\small{ \begin{array}{rcll} f(x) &=& 0,5x^2 -3 \qquad & | \qquad \text{ Nullstelle } f(x) = 0\\ 0,5x^2 -3 &=& 0 \qquad & | \qquad + 3\\ 0,5x^2 &=& 3 \qquad & | \qquad :0,5\\ x^2 &=& 6 \qquad & | \qquad \pm\sqrt{}\\ x &=& \pm \sqrt{6} \\\\ x_1 &=& +\sqrt{6} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2.44948974278} \\\\ x_2 &=&-\sqrt{6} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-2.44948974278} \end{array} }\)

e\(^x\)x = x\(^\pi\), what is x?

\(\small{ \begin{array}{rcll} e^x &=& x^{\pi} \qquad & | \qquad \ln{}\\ x\ln{(e)} &=& \pi \cdot \ln{(x)} \\ \frac{x}{\pi} \cdot \ln{(e)}&=& \ln{(x)} \\ \ln{( e^{\frac{x}{\pi}} )} &=& \ln{(x)} \qquad & | \qquad e^{}\\ e^{ \ln{( e^{\frac{x}{\pi}} )} } &=& e^{ \ln{(x)} }\\ e^{\frac{x}{\pi}} &=& x \\ x \cdot e^{-\frac{x}{\pi}} &=& 1 \qquad & | \qquad : -\pi\\ -\frac{x}{\pi} \cdot e^{-\frac{x}{\pi}} &=& -\frac{1}{\pi}\\ \\ \hline \end{array} }\\ \small{ \begin{array}{rcll} \text{Lambert W-Function(W) or ProductLog or Omega-Function } w\cdot e^w = z \qquad w = W(z) \end{array} }\\ \small{ \begin{array}{rcl} -\frac{x}{\pi} \cdot e^{-\frac{x}{\pi}} &=& -\frac{1}{\pi} \\ -\frac{x}{\pi} &=& W( -\frac{1}{\pi} ) \\ -x &=& \pi \cdot W( -\frac{1}{\pi} )\\ \mathbf{x} & \mathbf{=} & \mathbf{-\pi \cdot W( -\frac{1}{\pi} )}\\ \\ \hline \\ W( -\frac{1}{\pi} ) &=& -0.553827036644513 \\ x &=& -\pi \cdot (-0.553827036644513) \\ \mathbf{ x } & \mathbf{\approx} & \mathbf{1.73989894968181} \\ \end{array} }\\\)

wie kommt man bei x^2+(2,4)^2=39^2 auf X

\(\small{ \begin{array}{rcll} x^2+2,4^2 &=& 39^2 \qquad & | \qquad -2,4^2\\ x^2 &=& 39^2 -2,4^2 \\ x^2 &=& 1521-5,76 \\ x^2 &=& 1515,24 \qquad & | \qquad \pm\sqrt{}\\ x &=& \pm\sqrt{1515,24} \\\\ x_1 & = & +\sqrt{1515,24}\\ \mathbf{x_1} &\mathbf{=} & \mathbf{38.9260837999} \\\\ x_2 &=& -\sqrt{1515,24} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-38.9260837999} \end{array} } \)

2 to the power of 3 over 2

\( \small{ \begin{array}{rcl} 2^{ \frac 32 } \\ &=& 2^{ \frac{1+2}{2} } \\ &=& 2^{ \frac{1}{2}+\frac{2}{2} } \\ &=& 2^{ \frac{1}{2}+ 1 } \\ &=& 2^{ \frac{1}{2} }\cdot 2^1 \\ &=& 2^{ \frac{1}{2} }\cdot 2 \\ &=& \sqrt{2}\cdot 2 \\ &=& 2 \cdot \sqrt{2}\\ &=& 2.82842712475 \end{array} }\)

1-(x+3)\(\ge\)4-2x

\( \small{ \begin{array}{lrcll} & 1-(x+3) &\ge& 4-2x \\ \\ \hline \\ (1) & 1-(x+3) & = & 4-2x \\ & 1 - x - 3 &=& 4 - 2x \\ & -2 - x &=& 4 - 2x \qquad & | \qquad + 2x\\ & -2 + x &=& 4 \qquad & | \qquad +2\\ & \mathbf{x} & \mathbf{=} & \mathbf{6 }\\ \\ \hline \\ (2) & 1-(x+3) & > & 4-2x \\ & 1 - x - 3 & > & 4 - 2x \\ & -2 - x & > & 4 - 2x \qquad & | \qquad + 2x\\ & -2 + x &> & 4 \qquad & | \qquad +2\\ & \mathbf{x} & \mathbf{>} & \mathbf{6 }\\ \\ \hline \\ (1) + ( 2) & x = 6 ~ \text{ and } ~ x > 6 \\ & \mathbf{ x \ge 6 } \end{array} }\)

Kann ich bei dieser Formel ((2ax+8ab-2ay)/(2ax-2ay)) nicht einfach 2ax und 2ay wegkürzen ?

\(\small{ \begin{array}{rcr} \frac{ 2ax+8ab-2ay } { 2ax-2ay } \\ &=& \frac{ 2ax-2ay +8ab } { 2ax-2ay } \\ &=& \frac{ (2ax-2ay)+8ab } { 2ax-2ay } \\ &=& \frac{ 2ax-2ay } { 2ax-2ay } + \frac{ 8ab } { 2ax-2ay } \\ &=& 1 + \frac{ 8ab } { 2ax-2ay } \\ &=& 1 + \frac{ 8ab } { 2a(x-y) } \\ &=& 1 + \frac{ 4b } { x-y } \\ \end{array} }\)

was sind dreieckszahlen?

1
1+2=3
(1+2)+3=6
(1+2+3)+4=10
(1+2+3+4)+5=15
...

oder

\(\begin{array}{rcr} 1 & \qquad = 1\\\\ 1 \\ 11& \qquad = 3\\\\ 1 \\ 11 \\ 111& \qquad = 6\\\\ 1 \\ 11 \\ 111\\ 1111& \qquad = 10\\\\ 1 \\ 11 \\ 111\\ 1111\\ 11111& \qquad = 15\\\\ 1 \\ 11 \\ 111\\ 1111\\ 11111\\ 111111& \qquad = 21\\\\ \dots \end{array}\)

A number divided by 25 leaves remainder 9 . what is that number?

\(x\equiv 9 \pmod {25} \\ \text{or}\\ x-9 = z\cdot 25\\ x = 9+z\cdot 25 \qquad z \in Z\\\\ \mathbf{ x = \dots,-41, -16, 9, 34, 59, 84,\dots }\)