2.5-bo'lim : Kvadrat tenglamalar - I qism
10. Quyidagi tenglamani yechish uchun ko'paytuvchilarga ajratishdan foydalaning.
\[\frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\]Yechim
Bu ratsional ifodalarni o'z ichiga olgan tenglama, shuning uchun biz bilamizki, birinchi qadam LCD displeyga ko'paytirish orqali maxrajni tozalashdir, bu holda \(w + 2\). Shuni ham yodda tutingki, biz endi \(w = - 2\) dan qochishimiz kerakligini bilamiz, shuning uchun biz nolga bo'linmaymiz.
LCD displeyga ko'paytirish va ba'zi bir oddiy soddalashtirishni amalga oshirish,
\[\begin{align*}\left( {w + 2} \right)\left( {\frac{{{w^2} - 10}}{{w + 2}} + w - 4} \right) & = \left( {w - 3} \right)\left( {w + 2} \right)\\ {w^2} - 10 + \left( {w - 4} \right)\left( {w + 2} \right) & = \left( {w - 3} \right)\left( {w + 2} \right)\\ {w^2} - 10 + {w^2} - 2w - 8 & = {w^2} - w - 6\\ {w^2} - w - 12 & = 0\end{align*}\]2-qadam
Endi kvadratni koeffitsient qilishimiz mumkin,
\[\left( {w - 4} \right)\left( {w + 3} \right) = 0\]Nolinchi omil xususiyati endi bizga aytadi:
\[\begin{array}{*{20}{c}}{w - 4 = 0}\\{w = 4}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{w + 3 = 0}\\{w = - 3}\end{array}\]Shuning uchun ikkita yechim: \(\require{bbox} \bbox[2pt,border:1px solid black]{{w = 4\,\,{\mbox{and }}w = - 3}}\).
Shuni ham yodda tutingki, ularning hech biri \(w = - 2\) emas, chunki biz nolga bo'linmasligimizni bilamiz. Ratsional ifodalarni o'z ichiga olgan tenglamalar uchun yechim jarayonining bu muhim qismini unutmang!