5.5-bo'lim : qisman kasrlar
7. Quyidagi ifodalarning har birining qisman kasr parchalanishini aniqlang.
\[\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}\]Yechim
Birinchi qadam qisman fraksiyaning parchalanish shaklini aniqlashdir. Ushbu muammo uchun qisman kasr parchalanishi,
\[\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{2x + 3}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}}\]2-qadam
Ushbu ifoda uchun LCD displey \(\left( {2x + 3} \right){\left( {x - 2} \right)^2}\). Shartlarni qo'shish zaxira nusxasini beradi,
\[\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{{A{{\left( {x - 2} \right)}^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}\]3-qadam
Numeratorlarni teng belgilash,
\[4{x^2} - 22x + 7 = A{\left( {x - 2} \right)^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)\]4-qadam
Ushbu muammo uchun biz uchta doimiydan faqat ikkitasini aniqlash uchun \(x\) ning "yaxshi" qiymatlarini tanlashimiz mumkin. Mana o'sha ish.
\[\begin{array}{l}{\displaystyle x = - \frac{3}{2}:}\\{x = 2:}\end{array}\hspace{0.25in}\begin{aligned}49 & = A{{\left( { - \frac{7}{2}} \right)}^2} = \frac{{49}}{4}A\\ - 21 & = 7C\end{aligned}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{A = 4}\\{C = - 3}\end{array}\]5-qadam
Qolgan konstantani olish uchun biz \(x\) ning istalgan qiymatidan foydalanishimiz va uni oldingi bosqichda topilgan \(A\) va \(C\) qiymatlari bilan bir qatorda 3-bosqichdagi tenglamaga ulashimiz mumkin.
Biz \(x\) ning qaysi qiymatini tanlashimiz muhim emas, chunki u \(x = - \frac{3}{2}\) yoki \(x = 2\) emas, chunki biz avvalgi bosqichda ulardan foydalanganmiz. Bu erda g'oya, agar iloji bo'lsa, "katta" yoki "tartibsiz" raqamlarni yaratmaydigan \(x\) qiymatini tanlashdir. Yaxshi tanlovlar odatda \(x = 0\) yoki \(x = 1\) bo'ladi, agar ular avvalgi bosqichda ishlatilmagan bo'lsa.
Ushbu muammo uchun \(x = 0\) yaxshi tanlov bo'lib tuyuladi. Mana bu bosqich uchun ish.
\[\begin{align*}7 & = \left( 4 \right){\left( { - 2} \right)^2} + B\left( 3 \right)\left( { - 2} \right) + \left( { - 3} \right)\left( 3 \right)\\ 7 & = 7 - 6B\\ 0 & = - 6B\hspace{0.25in} \Rightarrow \hspace{0.25in}B = 0\end{align*}\]Bunday holda, konstantalardan biri nolga teng bo'ldi. Bu ba'zida sodir bo'ladi, lekin umuman olganda, bu sodir bo'lishini oldindan bilishning iloji yo'q, shuning uchun tashvishlanmang. Agar doimiylardan biri nolga teng bo'lsa, biz buni ishni bajarganimizda aniqlaymiz.
6-qadam
Bunda qisman fraksiya parchalanishi,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{4}{{2x + 3}} - \frac{3}{{{{\left( {x - 2} \right)}^2}}}}}\]