Asosiy tarkibga o'tish

2.13-bo'lim: Ratsional tengsizliklar

Ushbu bo'limda biz ratsional ifodalarni o'z ichiga olgan tengsizliklarni yechamiz. Ratsional tengsizliklarni yechish jarayoni echish jarayoni bilan deyarli bir xil polinom tengsizliklari bir nechta kichik farqlar bilan.

Keling, to'g'ridan-to'g'ri ba'zi misollarga o'taylik.

1-misol \(\displaystyle \frac{{x + 1}}{{x - 5}} \le 0\) ni yeching.
Yechim

Bularni echishga kirishishdan oldin shuni ta'kidlashimiz kerakki, bular biz ratsional ifodalarni o'z ichiga olgan tenglamalarni yechganimiz kabi EMAS. Tenglamalar bilan biz har doim qilgan birinchi narsa eng kichik umumiy maxrajga ko'paytirish orqali maxrajlarni aniqlash edi. Biroq, bular bilan ishlamaydi.

Biz \(x\) qiymatini bilmaganimiz uchun ikkala tomonni \(x\) ni o'z ichiga olgan har qanday narsaga ko'paytira olmaymiz. Eslatib o'tamiz, agar biz tengsizlikning ikkala tomonini manfiy songa ko'paytirsak, tengsizlikning yo'nalishini o'zgartirishimiz kerak bo'ladi. Biroq, biz \(x\) qiymatini bilmaganimiz uchun biz maxrajning ijobiy yoki salbiy ekanligini bilmaymiz va shuning uchun biz tengsizlik yo'nalishini o'zgartirishimiz kerakmi yoki yo'qligini bilmaymiz. Haqiqatan ham, vaziyatni yomonlashtirish uchun, maxraj yechimdagi \(x\) qiymatlari uchun ham ijobiy, ham salbiy bo'ladi va bu haqiqiy muammolarni keltirib chiqaradi.

Demak, ratsional ifodani tengsizlikda qoldirishimiz kerak.

Endi bu erda asosiy jarayon polinom tengsizliklari bilan bir xil. Birinchi qadam bir tomonda nol olish va boshqa tomonni yagona ratsional tengsizlik sifatida yozishdir. Bu biz uchun bu erda allaqachon qilingan.

Keyingi qadam imkon qadar pay va maxrajni koeffitsient qilishdir. Shunga qaramay, bu biz uchun bu holatda allaqachon qilingan.

Keyingi qadam, hisoblagich ham, maxraj ham nolga teng ekanligini aniqlashdir. Bunday holda, bu qiymatlar.

\[{\mbox{surat : }}x = - 1\hspace{0.5in}{\mbox{maxraj : }}x = 5\]

Endi bu raqamlar bizga bir nechta sabablarga ko'ra kerak. Birinchidan, xuddi polinom tengsizliklari kabi, bu ratsional ifoda bo'lgan yagona raqamlar mumkin belgisini o'zgartirish. Shunday qilib, biz polinom tengsizliklari kabi sinov nuqtalarini tanlash uchun diapazonlarni aniqlash uchun ushbu nuqtalardan foydalangan holda raqamlar chizig'ini quramiz.

\(x\) qiymatiga ehtiyojning yana bir sababi bor, bu esa maxrajni nolga tenglashtiradi. Bu erda nima bo'lishidan qat'i nazar, bizda ratsional ifoda bor va bu biz nolga bo'linishdan qochishimiz kerakligini anglatadi va shuning uchun maxrajning qaerda nol ekanligini bilish bizga buning oldini olish uchun \(x\) qiymatlarini beradi.

Mana bu tengsizlik uchun sonlar qatori.

Basic number line with scale in the range from -3 < x < 7 and divided into three ranges by vertical dashed lines at x=-1 and x=5. In the range x < -1 the rational expression is (-1)/(-7)>0, so positive, at the test point of x=-2. In the range -1 < x < 5 the rational expression is (1)/(-5)<0, so negative, at the test point of x=0. In the range x > 5 the rational expression is (7)/(1)>0, so positive, at the test point of x=6.

Shunday qilib, bizga ratsional ifodani salbiy qiladigan mintaqalar kerak. Bu o'rta mintaqani anglatadi. Bundan tashqari, biz tengsizlikda "yoki teng" qismga ega bo'lganimiz sababli, biz tengsizlik nolga teng bo'lgan joyni ham kiritishimiz kerak, shuning uchun biz \(x = - 1\) ni qo'shamiz. E'tibor bering, biz \(x = 5\) dan qochishimiz kerak, chunki bu nolga bo'linishni beradi.

Bu tengsizlikning yechimi:

\[ - 1 \le x < 5\hspace{0.25in}\left[ { - 1,5} \right)\]
2-misol \(\displaystyle \frac{{{x^2} + 4x + 3}}{{x - 1}} > 0\) ni yeching.
Yechim

Bizda bir tomonda nol bor, shuning uchun keling, birinchi navbatda hisobni ko'paytiramiz va pay va maxraj qaerda nolga teng ekanligini aniqlaymiz.

\[\frac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{x - 1}} > 0\] \[{\mbox{surat : }}x = - 1,\,\,\,\,x = - 3\hspace{0.5in}{\mbox{maxraj : }}x = 1\]

Mana bu raqam qatori.

Basic number line with scale in the range from -4 < x < 2 and divided into four ranges by vertical dashed lines at x=-3, x=-1 and x=1. In the range x < -3 the rational expression is (-3)(-1)/(-5)<0, so negative, at the test point of x=-4. In the range -3 < x < -1 the rational expression is (1)(-1)/(-3)>0, so positive, at the test point of x=-2. In the range -1 < x < 1 the rational expression is (3)(1)/(-1)<0, so negative, at the test point of x=0. In the range x > 1 the rational expression is (5)(3)/(1)>0, so positive, at the test point of x=2.

Muammoda biz \(x\) qiymatlaridan keyin turamiz, ular tengsizlikni qat'iy ijobiy qiladi va shuning uchun ikkinchi va to'rtinchi mintaqaga o'xshaydi va biz bu erda hech qanday so'nggi nuqtani kiritmaymiz. Yechim shundan iboratki,

\[\begin{array}{c} - 3 < x < - 1\hspace{0.25in}\text{va}\hspace{0.25in}1 < x < \infty \\\left( { - 3, - 1} \right)\hspace{0.25in}\text{va}\hspace{0.25in}\,\,\,\,\left( {1,\infty } \right)\end{array}\]
3-misol \(\displaystyle \frac{{{x^2} - 16}}{{{{\left( {x - 1} \right)}^2}}} < 0\) ni yeching.
Yechim

Haqiqatan ham bu misol uchun juda ko'p narsa yo'q. Biz birinchi navbatda suratni koeffitsientga kiritishimiz kerak, keyin esa pay va maxrajning qayerda nolga teng ekanligini aniqlashimiz kerak.

\[\frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{{{\left( {x - 1} \right)}^2}}} < 0\] \[{\mbox{surat : }}x = - 4,\,\,\,\,x = 4\hspace{0.5in}{\mbox{maxraj : }}x = 1\]

Bu muammo uchun raqam chizig'i,

Basic number line with scale in the range from -6 < x < 6 and divided into four ranges by vertical dashed lines at x=-4, x=1 and x=4. In the range x < -4 the rational expression is (-9)(-1)/(36)>0, so positive, at the test point of x=-5. In the range -4 < x < 1 the rational expression is (-4)(4)/(1)<0, so negative, at the test point of x=0. In the range 1 < x < 4 the rational expression is (-2)(6)/(1)<0, so negative, at the test point of x=2. In the range x > 4 the rational expression is (1)(9)/(16)>0, so positive, at the test point of x=2.

Shunday qilib, polinom tengsizliklarida bo'lgani kabi, biz mintaqalar har doim bir-birini almashtirib turadi deb taxmin qila olmaymiz. Shuni ham yodda tutingki, o'rtadagi ikkita mintaqa mantiqiy ifodada salbiy qiymatlarni bersa-da, biz nolga bo'linmasligimiz uchun \(x = 1\) dan qochishimiz kerak. Bu shuni anglatadiki, biz javobni ikkita tengsizlik va/yoki interval sifatida yozishimiz kerak bo'ladi.

\[\begin{array}{c} - 4 < x < 1\hspace{0.25in}\text{va}\hspace{0.25in}1 < x < 4\\ \left( { - 4,1} \right)\hspace{0.25in}\text{va}\hspace{0.25in}\,\,\,\,\left( {1,4} \right)\end{array}\]

Yana bir bor shuni ta'kidlash kerakki, biz haqiqatan ham har bir mintaqani sinab ko'rishimiz kerak va mintaqalar bir-birini almashtiradi deb o'ylamasdan.

Keyinchalik, biz tengsizlikning bir tomonida nolga ega bo'lmagan ba'zi misollarni ko'rib chiqishimiz kerak.

4-misol \(\displaystyle \frac{{3x + 1}}{{x + 4}} \ge 1\) ni yeching.
Yechim

Bu erda qilishimiz kerak bo'lgan birinchi narsa - ikkala tomondan 1 ni ayirish va keyin hamma narsani bitta ratsional ifodaga olish.

\[\begin{align*}\frac{{3x + 1}}{{x + 4}} - 1 & \ge 0\\ \frac{{3x + 1}}{{x + 4}} - \frac{{x + 4}}{{x + 4}} & \ge 0\\ \frac{{3x + 1 - \left( {x + 4} \right)}}{{x + 4}} & \ge 0\\ \frac{{2x - 3}}{{x + 4}} & \ge 0\end{align*}\]

Bu holatda faktoring yo'q, shuning uchun biz to'g'ridan-to'g'ri pay va maxraj nolga teng ekanligini aniqlashga o'tamiz.

\[{\mbox{surat : }}x = \frac{3}{2}\hspace{0.5in}{\mbox{maxraj : }}x = - 4\]

Mana bu muammo uchun raqam chizig'i.

Basic number line with scale in the range from -6 < x < 4 and divided into three ranges by vertical dashed lines at x=-4 and x=1.5. In the range x < -4 the rational expression is (-13)/(-1)>0, so positive, at the test point of x=-5. In the range -4 < x < 1.5 the rational expression is (-3)/(4)<0, so negative, at the test point of x=0. In the range x > 1.5 the rational expression is (1)/(6)>0, so positive, at the test point of x=2.

OK, biz \(x\) ning ratsional ifodada musbat va/yoki nol qiymatlarini istaymiz. Bu tashqi ikki mintaqaga, shuningdek \(x = \frac{3}{2}\) ga o'xshaydi. Birinchi misolda bo'lgani kabi biz \(x = - 4\) dan qochishimiz kerak, chunki bu nolga bo'linish xatosini beradi.

Bu muammoning yechimi shundan iboratki,

\[\begin{array}{c} \displaystyle - \infty < x < - 4\hspace{0.25in}\text{va}\hspace{0.25in}\frac{3}{2} \le x < \infty \\ \displaystyle \left( { - \infty , - 4} \right)\hspace{0.25in}\text{va}\hspace{0.25in}\,\,\,\left[ {\frac{3}{2},\infty } \right)\end{array}\]
5-misol \(\displaystyle \frac{{x - 8}}{x} \le 3 - x\) ni yeching.
Yechim

Shunday qilib, yana birinchi narsa bir tomondan nolni olish va keyin hamma narsani bitta ratsional ifodaga olishdir.

\[\begin{align*}\frac{{x - 8}}{x} + x - 3 & \le 0\\ \frac{{x - 8}}{x} + \frac{{x\left( {x - 3} \right)}}{x} & \le 0\\ \frac{{x - 8 + {x^2} - 3x}}{x} & \le 0\\ \frac{{{x^2} - 2x - 8}}{x} & \le 0\\ \frac{{\left( {x - 4} \right)\left( {x + 2} \right)}}{x} & \le 0\end{align*}\]

Biz yuqoridagi payni ham faktorlarga ajratdik, shuning uchun endi pay va maxraj qayerda nolga teng ekanligini aniqlashimiz mumkin.

\[{\mbox{surat : }}x = - 2,\,\,\,\,x = 4\hspace{0.5in}{\mbox{maxraj : }}x = 0\]

Mana bu muammo uchun raqam chizig'i.

Basic number line with scale in the range from -4 < x < 4 and divided into four ranges by vertical dashed lines at x=-2, x=0 and x=4. In the range x < -2 the rational expression is (-1)(-7)/(-3)<0, so negative, at the test point of x=-3. In the range -2 < x < 0 the rational expression is (1)(-5)/(-1)>0, so positive, at the test point of x=-1. In the range 0 < x < 4 the rational expression is (-3)(3)/(1)<0, so negative, at the test point of x=1. In the range x > 4 the rational expression is (7)(1)/(5)>0, so positive, at the test point of x=5.

Bu tengsizlikning yechimi,

\[\begin{array}{c} - \infty < x \le - 2\hspace{0.25in}\text{va}\hspace{0.25in}0 < x \le 4\\ \left( { - \infty , - 2} \right]\hspace{0.25in}\text{va}\hspace{0.25in}\,\,\,\,\left( {0,4} \right]\end{array}\]