2.5-bo'lim : Kvadrat tenglamalar - I qism
12. Quyidagi tenglamani yechish uchun ko'paytuvchilarga ajratishdan foydalaning.
\[x + 1 = \frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}\]Yechim
Bu ratsional ifodalarni o'z ichiga olgan tenglama, shuning uchun biz bilamizki, birinchi qadam LCD displeyga ko'paytirish orqali maxrajni tozalashdir, bu holda \(x + 5\). Shuni ham yodda tutingki, biz endi \(x = - 5\) dan qochishimiz kerakligini bilamiz, shuning uchun biz nolga bo'linmaymiz.
LCD displeyga ko'paytirish va ba'zi bir oddiy soddalashtirishni amalga oshirish,
\[\begin{align*}\left( {x + 5} \right)\left( {x + 1} \right) & = \left( {\frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}} \right)\left( {x + 5} \right)\\ \left( {x + 5} \right)\left( {x + 1} \right) & = \left( {\frac{{2x - 7}}{{x + 5}}} \right)\left( {x + 5} \right) - \left( {\frac{{5x + 8}}{{x + 5}}} \right)\left( {x + 5} \right)\\ \left( {x + 5} \right)\left( {x + 1} \right) & = 2x - 7 - \left( {5x + 8} \right)\\ {x^2} + 6x + 5 & = 2x - 7 - 5x - 8\\ {x^2} + 9x + 20 & = 0\end{align*}\]2-qadam
Endi kvadratni koeffitsient qilishimiz mumkin,
\[\left( {x + 4} \right)\left( {x + 5} \right) = 0\]Nolinchi omil xususiyati endi bizga aytadi:
\[\begin{array}{*{20}{c}}{x + 4 = 0}\\{x = - 4}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{x + 5 = 0}\\{x = - 5}\end{array}\]E'tibor bering, biz ikkinchi potentsial yechimdan foydalana olmaymiz, chunki bu bizga nolga bo'linish imkonini beradi! Shuning uchun yagona yechim: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 4}}\).
Ratsional ifodalarga ega bo'lgan tenglamalar bilan ishlashda siz hech qanday potentsial echim bilan nolga bo'linmasligingizni tekshirishni unutmang! Bu holatda ko'rganimizdek, agar biz tekshirmaganimizda \(x\) qiymatini olgan bo'lardik, bu yechim bo'lib tuyuldi, lekin aslida nolga bo'linish tufayli emas edi.