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\ANG{
Vereinfache den Term \newline
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\begin{tabular}[h]{lll}
(a) $3 \sqrt a + 5 \sqrt a \qquad$
& (b) $7 \sqrt b - 12 \sqrt b \qquad$ 
& (c) $7 \sqrt a \cdot 3 \sqrt b \qquad$ \\
(d) $3 \sqrt a \cdot 5 \sqrt a \qquad$ 
& (e) $\D\frac {16 \sqrt a}{2 \sqrt b}\qquad$
& (f) $\D\frac {2 \sqrt b}{\frac 1 3 \sqrt b} \qquad$
\end{tabular}
}
\LSG{ 
Vereinfache den Term \\
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\begin{tabular}[h]{ll}
(a) $3 \sqrt a + 5 \sqrt a = \left(3+5 \right) \sqrt a = 8 \sqrt a$
& (b) $7 \sqrt b - 12 \sqrt b = \left( 7 - 12 \right) \sqrt b = -5 \sqrt b$ \\
(c) $7 \sqrt a \cdot 3 \sqrt b = 7 \cdot 3 \cdot \sqrt a \cdot \sqrt b = 21 \sqrt {ab}$ 
& (d) $3 \sqrt a \cdot 5 \sqrt a = 15 \cdot \sqrt a \cdot \sqrt a = 15 a$  \\
(e) $\D\frac {16 \sqrt a}{2 \sqrt b} = \D\frac {8 \sqrt a}{\sqrt b} = 8 \sqrt{\frac a b}$
& (f) $\D\frac {2 \sqrt b}{\frac 1 3 \sqrt b} = 2 \cdot \frac 3 1 = 6$
\end{tabular}
}
