\( \left. \begin{array} { l } { a , b \in N } \\ { ( \frac { 2 a + 1 } { b - 3 } ) \in Z } \\ { ( \frac { b - 3 } { 2 a + 1 } ) \in Z } \\ { \Rightarrow \min ( a + b ) = ? } \end{array} \right. \)

(8) If \( L^{2} \) and \( L \) are the two roots of the equation : \( x^{2}+a x+8=0 \), then \( a=\ldots \ldots \ldots \) \( \begin{array}{llll}\text { (a) }-6 & \text { (b) } 6 & \text { (c) } 4 & \text { (d) } 8\end{array} \)

Simplify the following: (1) \( \frac{x-1}{x^{2}+3 x+2} \div\left[\frac{x-3}{x+1}-\frac{x-4}{x+2}\right] \) (2) \( \frac{\frac{6 x+5}{x-5}}{\frac{3 x+20}{x-5}+2} \) (3) \( \frac{a^{2}+\left(1-a^{2}\right)}{(1+a) a} \times \frac{(1+a) a}{a^{2}-\left(1-a^{2}\right)} \)

Simplify. \( 6 v+3 v \)

Evaluate the expression when \( a=6 \) and \( b=27 \). \[ b-4 a \] \( \square \) \begin{tabular}{ccc}\hline\( \square \) & 5 \\ \hline\end{tabular}

Example Solve the equation \( 2 x^{2}+4 x-10=0 \) with error precision 0.05