Реши уравнения

а) \(\dfrac{(3m-1)(m+1)}{5}+\dfrac{(2m-1)^2}{3}=5+\dfrac{3m-m^2}{2}\) ;

б) \(\dfrac{(p+3)^2}{4}-\dfrac{(p+2)(p-3)}{6}=(2p-1)^2+4\) .

Решение.

Запиши корни в порядке возрастания.

\(\frac{(\mathrm{p}+3)^{2}}{4}-\frac{(\mathrm{p}+2)(\mathrm{p}-3)}{6}=(2 \mathrm{p}-1)^{2}+4\) ;

\(\frac{\left\{\mathrm{p}^{2}+6 \mathrm{p}+9\right\}}{4}-\frac{\mathrm{p}^{2}-\mathrm{p}-6}{6}=\left\{4 \mathrm{p}^{2}-4 \mathrm{p}+1\right\}+4\) ;

\(\frac{\{3\} \cdot \mathrm{p}^{2}+\{18\} \cdot \mathrm{p}+\{27\}}{12}-\frac{\{2\} \cdot \mathrm{p}^{2}-\{2\} \cdot \mathrm{p}-\{12\}}{12}=\frac{\{48\} \cdot \mathrm{p}^{2}-\{48\} \cdot \mathrm{p}+\{60\}}{12}\) ;

\(\{3\} \cdot \mathrm{p}^{2}+\{18\} \cdot \mathrm{p}+\{27\}-\{2\} \cdot \mathrm{p}^{2}\{+/-\}\{2\} \cdot \mathrm{p}\{+/-\}\{12\}\{+/-\}\{48\} \cdot \mathrm{p}^{2}\{+/-\}\{48\} \mathrm{p}\{+/-\}\{60\} \\=0\) ;

\(\{47\} \cdot \mathrm{p}^{2}-\{68\} \cdot \mathrm{p}+\{21\}=0\) ;

\(\mathrm{p}\_{1}=\{21\}, \mathrm{p}\_{2}=\{47\}\) .