heureka

\(\small{ \begin{array}{lrcl} (v) & \sqrt{3x^3y^2} \cdot \sqrt[3]{3x^4y^{-5}} \cdot \sqrt[6]{3xy^4} =\\ &&=& 3^{\frac{1}{2}}x^{ \frac{3}{2} } y^{ \frac{2}{2} } \cdot 3^{\frac{1}{3}}x^{ \frac{4}{3} } y^{ -\frac{5}{3} } \cdot 3^{\frac{1}{6}}x^{ \frac{1}{6} } y^{ \frac{4}{6} } \\ &&=& 3^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}} \cdot x^{ \frac{3}{2}+\frac{4}{3}+ \frac{1}{6} } \cdot y^{ \frac{2}{2}-\frac{5}{3}+ \frac{4}{6}} \\ &&=& 3^{\frac{3}{6}+\frac{2}{6}+\frac{1}{6}} \cdot x^{ \frac{3}{2}+\frac{4}{3}+ \frac{1}{6} } \cdot y^{ \frac{2}{2}-\frac{5}{3}+ \frac{4}{6}} \\ &&=& 3^{1} \cdot x^{ \frac{3}{2}+\frac{4}{3}+ \frac{1}{6} } \cdot y^{ \frac{2}{2}-\frac{5}{3}+ \frac{4}{6}} \\ &&=& 3^{1} \cdot x^{ \frac{3\cdot 3}{6}+\frac{4\cdot 2}{6}+ \frac{1}{6} } \cdot y^{ \frac{2\cdot 3}{6}-\frac{5\cdot 2}{6}+ \frac{4}{6}} \\ &&=& 3 \cdot x^{ \frac{9}{6}+\frac{8}{6}+ \frac{1}{6} } \cdot y^{ \frac{6}{6}-\frac{10}{6}+\frac{4}{6}} \\ &&=& 3 \cdot x^{ \frac{18}{6} } \cdot y^{ 0 } \\ &&=& 3 \cdot x^{ 3 } \cdot y^{ 0 } \\ &&=& 3 \cdot x^{ 3 } \cdot 1 \\ &&=& 3 x^3 \end{array} }\)

\(\small{ \begin{array}{lrcl} (t) & ( 3x^3y^5 )^{-3} \cdot ( 3x^{-2}y^{-2} )^{-2} \cdot ( 3xy^2 )^6 \\ &&=& 3^{-3} x^{3\cdot(-3)} y^{5\cdot(-3)} \cdot 3^{-2} x^{(-2)\cdot(-2)} y^{(-2)\cdot(-2)} \cdot 3^{6} x^{1\cdot 6} y^{2\cdot 6} \\ &&=& 3^{-3} x^{-9} y^{-15} \cdot 3^{-2} x^{4} y^{4} \cdot 3^{6} x^{6} y^{12} \\ &&=& 3^{-3-2+6} \cdot x^{-9+4+6} \cdot y^{-15+4+12} \\ &&=& 3^{1}\cdot x^{1}\cdot y^{1}\\ &&=& 3\cdot x\cdot y\\ \end{array} }\)

\(\small{ \begin{array}{lrcl} (r) & \dfrac { ( \sqrt[3]{x^5} \cdot \sqrt[5]{y^3} )^{-2} } { ( \sqrt[3]{x^{-2}} \cdot y^{-\frac{6}{25} } )^{5} }=\\\\ &&=& \dfrac { ( x^\frac{5}{3} \cdot y^\frac{3}{5} )^{-2} } { ( x^{-\frac{2}{3}} \cdot y^{-\frac{6}{25} } )^{5} }\\\\ &&=& \dfrac { x^{\frac{5\cdot(-2)}{3}} \cdot y^\frac{3\cdot(-2)}{5} } { x^{-\frac{2\cdot 5}{3}} \cdot y^{-\frac{6\cdot 5}{25} } }\\\\ &&=& \dfrac { x^{-\frac{10}{3}} \cdot y^{-\frac{6}{5} } } { x^{-\frac{10}{3}} \cdot y^{-\frac{6}{5} } }\\\\ &&=& x^{-\frac{10}{3}} \cdot x^{ \frac{10}{3}} \cdot y^{-\frac{6}{5} } \cdot y^{ \frac{6}{5} } \\ &&=& x^{-\frac{10}{3}+\frac{10}{3}} \cdot y^{-\frac{6}{5} + \frac{6}{5}} \\ &&=& x^{0} \cdot y^{0} \\ &&=& 1 \cdot 1 \\ &&=& 1 \end{array} }\)

\(\small{ \begin{array}{lrcl} (o) & \sqrt[3]{2^4} \cdot \sqrt[4]{2^3} \cdot \sqrt[12]{2^{-1} } =\\ &&=& 2^{\frac43}\cdot 2^{\frac34}\cdot 2^{-\frac{1}{12}} \\ &&=& 2^{\frac43+\frac34-\frac{1}{12}} \\ &&=& 2^{\frac{4\cdot4}{12}+\frac{3\cdot3}{12}-\frac{1}{12}} \\ &&=& 2^{\frac{16}{12}+\frac{9}{12}-\frac{1}{12}} \\ &&=& 2^{\frac{24}{12}} \\ &&=& 2^{2} \\ &&=& 4 \\ \end{array} }\)

\(\small{ \begin{array}{lrcl} (n) & ( 16x^8y^4 )^{0,5} \cdot ( 4x^4y )^{-1} \\ &&=& 16^{0,5}\cdot x^{8\cdot 0,5} \cdot y^{4 \cdot 0,5} \cdot 4^{-1}\cdot x^{4\cdot (-1)} \cdot y^{-1}\\ &&=& 16^{0,5}\cdot x^{4} \cdot y^{2} \cdot 4^{-1}\cdot x^{-4} \cdot y^{-1}\\ &&=& \sqrt{16}\cdot x^{4} \cdot y^{2} \cdot 4^{-1}\cdot x^{-4} \cdot y^{-1}\\ &&=& 4\cdot x^{4} \cdot y^{2} \cdot 4^{-1}\cdot x^{-4} \cdot y^{-1}\\ &&=& 4\cdot 4^{-1}\cdot x^{4}\cdot x^{-4}\cdot y^{2} \cdot y^{-1}\\ &&=& 4^{1-1}\cdot x^{4-4}\cdot y^{2-1} \\ &&=& 4^{0}\cdot x^{0}\cdot y^{1} \\ &&=& 1 \cdot 1 \cdot y\\ &&=& y \end{array} } \)

\(\small{ \begin{array}{lrcl} (i) & \dfrac{ x^{\frac13} \cdot y^{\frac{1}{8}} \cdot x^{-\frac{4}{3}} \cdot y^{\frac{1}{8}} } { x^{-0,4} \cdot y^{-0,25} \cdot x^{-0,6} \cdot y^{-0,5} \cdot x^{-1} }= \\ &&=& x^{\frac13} \cdot x^{-\frac{4}{3}} \cdot x^{0,4} \cdot x^{0,6} \cdot x^{1} \cdot y^{\frac{1}{8}} \cdot y^{\frac{1}{8}} \cdot y^{0,25} \cdot y^{0,5} \\ &&=& x^{\frac13-\frac{4}{3}+0,4+0,6+1} \cdot y^{\frac{1}{8}+\frac{1}{8}+0.25+0.5} \\ &&=& x^{\frac13-\frac{4}{3}+2} \cdot y^{\frac{1}{8}+\frac{1}{8}+0.75} \\ &&=& x^{-\frac33+2} \cdot y^{\frac{1}{8}+\frac{1}{8}+0.75} \\ &&=& x^{-1+2} \cdot y^{\frac{1}{8}+\frac{1}{8}+0.75} \\ &&=& x^{1} \cdot y^{\frac{1}{8}+\frac{1}{8}+0.75} \\ &&=& x^{1} \cdot y^{\frac{2}{8}+0.75} \\ &&=& x^{1} \cdot y^{\frac{1}{4}+0.75} \\ &&=& x^{1} \cdot y^{0,25+0.75} \\ &&=& x^{1} \cdot y^{1} \\ &&=& x \cdot y \\ \end{array} } \)

\(\small{ \begin{array}{lrcl} (h) & \dfrac{ 5^{\frac12} \cdot 5^{-\frac{2}{5}} \cdot 5^{\frac{2}{3}} \cdot 5^{-\frac{9}{10}} } { 5^{\frac65} \cdot 5^{-\frac{1}{2}} \cdot 5^{-\frac{5}{6}} \cdot 5^{-2} }= \\ &&=& 5^{\frac12} \cdot 5^{-\frac{2}{5}} \cdot 5^{\frac{2}{3}} \cdot 5^{-\frac{9}{10}} \cdot 5^{-\frac65} \cdot 5^{\frac{1}{2}} \cdot 5^{\frac{5}{6}} \cdot 5^{2}\\ &&=& 5^{\frac12-\frac{2}{5}+\frac{2}{3}-\frac{9}{10} -\frac65 + \frac{1}{2} + \frac{5}{6} + 2 } \\ &&=& 5^{\frac12+ \frac{1}{2} + 2+\frac{2}{3} -\frac{2}{5}-\frac{6}{5} + \frac{5}{6} -\frac{9}{10} } \\ &&=& 5^{1 + 2+\frac{2}{3}-\frac{8}{5}+ \frac{5}{6} -\frac{9}{10} } \\ &&=& 5^{3+\frac{2}{3}-\frac{8}{5}+ \frac{5}{6} -\frac{9}{10} } \\ &&=& 5^{3+\frac{2*10}{30}-\frac{8*6}{30}+ \frac{5*5}{30} -\frac{9*3}{30} } \\ &&=& 5^{3-\frac{30}{30} } \\ &&=& 5^{3-1} \\ &&=& 5^2\\ &&=& 25 \end{array} }\)

Factor the following polynomial: 36x^2−4

\(\small{\text{Formula Difference of Squares $~ \boxed{~ \textcolor[rgb]{1,0,0}{ a}^2 - \textcolor[rgb]{0,0,1}{ b}^2 = (\textcolor[rgb]{1,0,0}{ a} -\textcolor[rgb]{0,0,1}{ b})(\textcolor[rgb]{1,0,0}{ a} + \textcolor[rgb]{0,0,1}{b}) ~} $}}\\ \small{ \begin{array}{rcl} \\ 36x^2-4 & = & (6x)^2 -2^2\\ (\textcolor[rgb]{1,0,0}{6x})^2 -\textcolor[rgb]{0,0,1}{2}^2 &=& (\textcolor[rgb]{1,0,0}{6x}-\textcolor[rgb]{0,0,1}{2})(\textcolor[rgb]{1,0,0}{6x}+\textcolor[rgb]{0,0,1}{2})\\ \end{array} }\)

Wie kann ich bei einem Rechtwinkligen dreieck die seiten b und c und den Winkel berechnen wenn ich die Seiten a und Winkel Alpha habe?

1.)  \(\boxed{~\beta = 90^{\circ}- \alpha ~}\)

2.)

\(\small{ \begin{array}{rcl} \tan{\alpha} &=& \frac{a}{b} \\ \boxed{~b = \dfrac{a}{\tan{\alpha}}~} \end{array} }\)

3.)

\(\small{ \begin{array}{rcl} \boxed{~c = \sqrt{a^2+b^2}~} \\\\ \text{ oder }\\\\ \sin{(\alpha)} &=& \dfrac{a}{c} \\ \boxed{~c = \dfrac{a}{ \sin{(\alpha)}} ~}\\\\ \end{array} }\)

Solve for the indicated variable

2. cos2x+3cosx+2=0 ; x

\(\small{\text{$ \begin{array}{rcl} \cos^2{(x)} +3 \cos{(x)} +2 &=& 0 \qquad | \qquad \cos{(x)} = z \qquad x = \pm\arccos{(z)}\\ z^2 +3z+2 &=& 0 \\ z_{1,2} &=& \dfrac{-3\pm\sqrt{9-4\cdot 2} }{2} \\ z_{1,2} &=& \dfrac{-3\pm 1 }{2} \\ z_{1} &=& -1 \\ z_{2} &=& -2 \qquad \text{ no solution} \\\\ x_{1,2} &=& \pm \arccos{(z_1)} \\ x_{1,2} &=& \pm\arccos{(-1)} \\ x_1 &=& \pi \\ x_2 &=& -\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{\pi \pm 2k\pi } \qquad k \in N\\ \end{array} $}}\)