1.1-bo'lim: Ko'rsatkichlar
Quyidagilarning har birini iloji boricha soddalashtiring.
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\(2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}}\) Yechim
Bu muammolarning barchasi quyidagi xususiyatlarning bir yoki bir nechtasidan foydalanadi. \begin{massiv}{rclcrcl}{p^n}{p^m} & = & {p^{n + m}} & \hspace{0,5in} & \displaystyle \frac{{{p^n}}}{{{p^m}}} & = & {p^{n - m}} = \displaystyle \frac{1}{{\} {\left( {{p^n}} \right)^m} & = &{p^{nm}} & & {p^0} & = & 1\,{\mbox{, berilgan }}p \ne 0\\ {\ chap ( {pq} \ o'ng) ^ n} & = & {p ^ n} {q ^ n} & & {\ displaystyle \ chap ( {\ frac {p} {q}} \ o'ng) ^ n} & = & \ displaystyle \ frac {{{p ^ n}}} {{{q^n}}}\\ {p^{ - n}} & = & \displaystyle \frac{1}{{{p^n}}} & & \displaystyle \frac{1}{{{p^{ - n}}}} & = & {p^n}\\ \displaystyle {\left( {\frac{p}{q}} \right)^{ - n}} & = & \displaystyle {\left( {\frac{q}{p}} \right)^n} =\displaystyle \frac{{{q^n}}}{{{p^n}}} & & & & \end{massiv}Bu maxsus muammo faqat birinchi xususiyatdan foydalanadi.
\[2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}} = 2{x^{4 - 19}}{y^{ - 3}} + {y^{\frac{1}{3}\, - \,\frac{3}{4}}} = 2{x^{ - 15}}{y^{ - 3}} + {y^{ - \,\frac{5}{{12}}}}\]Esda tutingki, oxirgi ikki muddatdagi \(y\)-larni birlashtirib bo'lmaydi! Siz faqat mahsulot yoki ko'rsatkich bo'lgan atamalarni birlashtira olasiz. Bundan tashqari, bu Calculus kursida maqbul va ko'pincha afzal javob bo'lsa-da, algebra sinfi sizni salbiy ko'rsatkichlardan xalos bo'lishingizni xohlaydi. Bu holda sizning javobingiz bo'ladi.
\[2{x^4}{y^{ - 3}}{x^{ - 19}} + {y^{\frac{1}{3}}}{y^{ - \,\frac{3}{4}}} = 2{x^{ - 15}}{y^{ - 3}} + {y^{ - \,\frac{5}{{12}}}} = \frac{2}{{{x^{15}}{y^3}}} + \frac{1}{{{y^{\frac{5}{{12}}}}}}\]2 birinchi hadning suratida qoladi, chunki uning salbiy ko'rsatkichi yo'q.
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\({x^{\frac{3}{5}}}{x^2}{x^{ - \,\frac{1}{2}}}\) Yechim
\[{x^{\frac{3}{5}}}{x^2}{x^{ - \,\frac{1}{2}}} = {x^{\frac{3}{5}\, + \,2\, - \,\,\frac{1}{2}}} = {x^{\frac{6}{{10}}\, + \,\frac{{20}}{{10}}\, - \,\,\frac{5}{{10}}}} = {x^{\frac{{21}}{{10}}}}\]Bu yechim uchun ko'rsatkichlarni qo'shishdan boshqa narsa emas.
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\(\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}}\) Yechim
\[\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}} = \frac{{{x^{\frac{2}{3}}}}}{{2{x^5}}} = \frac{{{x^{\frac{2}{3}}}{x^{ - 5}}}}{2} = \frac{{{x^{ - \,\frac{{13}}{3}}}}}{2} = \frac{1}{2}{x^{ - \,\frac{{13}}{3}}}\]E'tibor bering, siz quyidagilarni ham qilishingiz mumkin edi (ehtimol osonroqdir ....).
\[\frac{{x{x^{ - \,\frac{1}{3}}}}}{{2{x^5}}} = \frac{{{x^{ - \,\frac{1}{3}}}}}{{2{x^4}}} = \frac{{{x^{ - \,\frac{1}{3}}}{x^{ - 4}}}}{2} = \frac{{{x^{ - \,\frac{{13}}{3}}}}}{2} = \frac{1}{2}{x^{ - \,\frac{{13}}{3}}}\]Ikkinchi holda, men har qanday soddalashtirishdan oldin \(x\) ni bekor qildim.
Ikkala holatda ham 2 maxrajda qoladi. Agar men 2 ning \(x\) bilan hisoblagichga kelishini xohlaganimda, maxrajda \({\left( {2x} \right)^5}\) dan foydalangan bo'lardim. Shunday qilib, qavsga qarang!
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\({\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}}\) Yechim
Ushbu muammoni hal qilishning bir necha yo'li mavjud. Men avval qavsning ichki qismini biroz soddalashtiraman. Shu bilan birga, men hamma narsada minus belgisidan xalos bo'lish uchun yuqoridagi oxirgi xususiyatdan foydalanaman.
\[{\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}} = {\left( {\frac{{x + y}}{{2{x^{ - \,\frac{6}{5}}}{y^6}}}} \right)^3}\]Endi ko'rsatkichni kiriting. Esda tutingki, har bir atama (shu jumladan 2) ko'rsatkichni olishi kerak.
\[{\left( {\frac{{2{x^{ - 2}}{x^{\frac{4}{5}}}{y^6}}}{{x + y}}} \right)^{ - 3}} = \frac{{{{\left( {x + y} \right)}^3}}}{{{2^3}{{\left( {{x^{ - \,\frac{6}{5}}}} \right)}^3}{{\left( {{y^6}} \right)}^3}}} = \frac{{{{\left( {x + y} \right)}^3}}}{{8{x^{ - \,\frac{{18}}{5}}}{y^{18}}}}\]Eslatib o'tamiz, \({\left( {x + y} \right)^3} \ne {x^3} + {y^3}\), shuning uchun siz bu bilan boshqa bora olmaysiz.
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\({\left( {\frac{{{x^{\frac{4}{7}}}{x^{\frac{9}{2}}}{x^{ - \,\frac{{10}}{3}}} - {x^2}{x^{ - 9}}{x^{\frac{1}{2}}}}}{{x + 1}}} \right)^0}\) Yechim
Buni bo'lishi kerak bo'lganidan qiyinlashtirmang. E'tibor bering, hamma narsa nol kuchga ko'tariladi, shuning uchun bu erda ishlatilishi kerak bo'lgan faqat bitta xususiyat mavjud.
\[{\left( {\frac{{{x^{\frac{4}{7}}}{x^{\frac{9}{2}}}{x^{ - \,\frac{{10}}{3}}} - {x^2}{x^{ - 9}}{x^{\frac{1}{2}}}}}{{x + 1}}} \right)^0} = 1\]