5.5-bo'lim : qisman kasrlar
6. Quyidagi ifodalarning har birining qisman kasr parchalanishini aniqlang.
\[\frac{{7{x^2} - 17x + 38}}{{\left( {x + 6} \right){{\left( {x - 1} \right)}^2}}}\]Yechim
Birinchi qadam qisman fraksiyaning parchalanish shaklini aniqlashdir. Ushbu muammo uchun qisman kasr parchalanishi,
\[\frac{{7{x^2} - 17x + 38}}{{\left( {x + 6} \right){{\left( {x - 1} \right)}^2}}} = \frac{A}{{x + 6}} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}}\]2-qadam
Ushbu ifoda uchun LCD displey \(\left( {x + 6} \right){\left( {x - 1} \right)^2}\). Shartlarni qo'shish zaxira nusxasini beradi,
\[\frac{{7{x^2} - 17x + 38}}{{\left( {x + 6} \right){{\left( {x - 1} \right)}^2}}} = \frac{{A{{\left( {x - 1} \right)}^2} + B\left( {x + 6} \right)\left( {x - 1} \right) + C\left( {x + 6} \right)}}{{\left( {x + 6} \right){{\left( {x - 1} \right)}^2}}}\]3-qadam
Numeratorlarni teng belgilash,
\[7{x^2} - 17x + 38 = A{\left( {x - 1} \right)^2} + B\left( {x + 6} \right)\left( {x - 1} \right) + C\left( {x + 6} \right)\]4-qadam
Ushbu muammo uchun biz uchta doimiydan faqat ikkitasini aniqlash uchun \(x\) ning "yaxshi" qiymatlarini tanlashimiz mumkin. Mana o'sha ish.
\[\begin{array}{l}{x = - 6:}\\{x = 1:}\end{array}\hspace{0.25in}\begin{aligned}392 & = A{{\left( { - 7} \right)}^2} = 49A \\28 & = 7C\end{aligned}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{A = 8}\\{C = 4}\end{array}\]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 \(x = - 6\) yoki \(x = 1\) 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 ko'pincha \(x = 0\) yoki \(x = 1\) (yuqorida aytib o'tilganidek, biz ushbu muammo uchun foydalana olmaymiz), agar ular avvalgi bosqichda ishlatilmagan bo'lsa.
Ushbu muammo uchun \(x = 0\) yaxshi tanlov bo'lib tuyuladi. Mana bu bosqich uchun ish.
\[\begin{align*}38 & = \left( 8 \right){\left( { - 1} \right)^2} + B\left( 6 \right)\left( { - 1} \right) + \left( 4 \right)\left( 6 \right)\\ 38 & = 32 - 6B\\ 6 & = - 6B\hspace{0.25in} \Rightarrow \hspace{0.25in}B = - 1\end{align*}\]6-qadam
Bunda qisman fraksiya parchalanishi,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{7{x^2} - 17x + 38}}{{\left( {x + 6} \right){{\left( {x - 1} \right)}^2}}} = \frac{8}{{x + 6}} - \frac{1}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}}}\]