2.14-bo'lim : Mutlaq qiymat tenglamalari
3. Quyidagi tenglamani yeching.
\[6u = \left| {1 + 3u} \right|\]Yechim
Mutlaq qiymat satrlaridan tashqaridagi miqdor ijobiy raqam emasligiga qaramay, biz birinchi ikkita muammoda ishlatgan jarayonni ishlata olmasligimizni anglatmaydi.
Eslatmalardagi formuladan foydalanish,
\[1 + 3u = - 6u\hspace{0.25in}{\mbox{or}}\hspace{0.25in}1 + 3u = 6u\]2-qadam
Endi bu chiziqli tenglamalarning har birini yechish,
\[\begin{align*}1 + 3u & = - 6u & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}1 + 3u & = 6u\\ 1 & = - 9u & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}1 & = 3u\\ u & = - \frac{1}{9} & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}u & = \frac{1}{3}\end{align*}\]3-qadam
Endi, mutlaq qiymat majmuasi tashqarisidagi miqdor ijobiy doimiy bo'lmaganligi sababli, oldingi bosqichda olgan javoblarimiz bilan ehtiyot bo'lishimiz kerak. Ehtimol, biri yoki ikkalasi ham asl tenglamaning yechimi emas. Shunday qilib, biz oldingi bosqichdagi mumkin bo'lgan echimlarning har birini dastlabki tenglamada tekshirish orqali tekshirishimiz kerak.
\[u = - \frac{1}{9}:\,\,\,\,\,\,\,\,6\left( { - \frac{1}{9}} \right)\mathop = \limits^? \left| {1 + 3\left( { - \frac{1}{9}} \right)} \right|\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\, - \frac{2}{3}\mathop = \limits^? \left| {\frac{2}{3}} \right|\,\,\,\,\,\,\, \to \,\,\,\,\,\,\, - \frac{2}{3} \ne \frac{2}{3}\hspace{0.25in}{\mbox{NOT}}{\mbox{ OK}}\] \[u = \frac{1}{3}:\,\,\,\,\,\,\,\,6\left( {\frac{1}{3}} \right)\mathop = \limits^? \left| {1 + 3\left( {\frac{1}{3}} \right)} \right|\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,2\mathop = \limits^? \left| 2 \right|\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,2 = 2\hspace{0.25in}{\mbox{OK}}\]Shunday qilib, dastlabki tenglamaning yagona yechimi: \(\require{bbox} \bbox[2pt,border:1px solid black]{{u = \frac{1}{3}}}\) .