to’g’ri chiziqda koordinatalar metodi

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1662926825.doc x b ab ab cd a a b d c ab ab ab ab ab ab cd cd ab c d a b ad ab cd ab cd a b c ab bc ac a b b c a c l ab 0 ¹ l ab l l ab l l ab ab l l ab l ab l ab o x ) ( x m o o o o m ом m x î í ì - = ом ом х x m ) ( x m 7 5 = - х x x x x ) ( 2 - n ) ( 12 m ) ( ), ( 2 2 1 1 x m x m 2 1 м м 1 2 x x - 1 2 2 1 x x m m - = 2 1 va , m m o 2 2 1 1 om …
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teng deb hisoblanadi. 2. yo’nalgan kesmalar ustida chiziqli amallar avvalo yo’nalgan kesmalarning tengligi tushunchasini aniqlaymiz. yo’nalgan kesmalarni o’q bo’ylab uning yunalishi va uzunligini o’zgartirmasdan siljitish mumkin deb hisoblaymiz. 5-ta’rif. agar o’q bo’ylab siljitilganda nol bo’lmagan ikki yo’nalgan kesmalarning boshidagi va oxiridagi nuqtalari ustma-ust tushsa, bu kesmalar o’zaro teng deyiladi. tasdiq. ikki yo’nalgan kesmaning o’zaro teng bo’lishi uchun ularning kattaliklari o’zaro teng bo’lishi zarur va yetarli. yo’nalgan kesmalarni qo’shish va haqiqiy songa ko’paytirish amallarini yunalgan kesmalar ustida chiziqli amallar deymiz. va yo’nalgan kesmalar berilgan bo’lsin. ularning yig’indisini topish uchun kesmaning boshini kesmaning oxiriga qo’yamiz (2-chizma) 2-chizma bu holda hosil bo’lgan kesma va yo’nalgan kesmalarning yig’indisi deyiladi va + kabi yoziladi. quyidagi teorema o’rinli. 1-teorema. yo’nalgan kesmalar yig’indisining kattaligi, qo’shiluvchi kesmalar kattaliklarining yig’indisiga teng. bu teoremadan quyidagi natija kelib chiqadi. natija. , va nuqtalar sonlar o’qi ustida qanday joylashmasin, , va yo’nalgan kesmalar kattaliklari + embed equation.3 = tenglikni qanoatlantiradi. bu asosiy …
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dinatalarini toping. yechish. berilgan nuqtalarni sonlar o’qida tasvirlaymiz (7-chizma). (-2) (5) (12) b (19) 7-chizma 1. shart bo’yicha c nuqta ad kesmani =1/2 nisbatda bo’ladi. (1) formulaga ko’ra x1 =-2, x2 =19, embed equation.3 deb olsak, nuqtaning koordinatasi: embed equation.3 yoki xc = 5 2. nuqta kesmani =2/1 nisbatda bo’ladi. (1) formulaga =-2, =19, =2 qiymatlarni qo’yib nuqtaning xd koordinatasini topamiz: xd =12 eslatma. ning musbat qiymatlari uchun nuqta va nuqtalar orasida yotadi ( 6-chizmaga qarang). bunda va kesmalar bir xil yo’nalgan bo’ladi). ning manfiy qiymatlari uchun nuqta kesmadan tashqarida yotadi (nima uchun ?). � embed equation.3 �� _1112722147.unknown _1112729136.unknown _1112733559.unknown _1112885477.unknown _1112885886.unknown _1142491359.unknown _1142491754.unknown _1166783355.unknown _1142491807.unknown _1142491378.unknown _1142491734.unknown _1142491338.unknown _1112886470.unknown _1112885591.unknown _1112885721.unknown _1112885883.unknown _1112885884.unknown _1112885747.unknown _1112885882.unknown _1112885678.unknown _1112885696.unknown _1112885612.unknown _1112885564.unknown _1112885359.unknown _1112885390.unknown _1112885456.unknown _1112885374.unknown _1112885312.unknown _1112885343.unknown _1112885148.unknown _1112729736.unknown _1112730074.unknown _1112730107.unknown _1112730141.unknown _1112730164.unknown _1112733558.unknown _1112730163.unknown _1112730123.unknown _1112730093.unknown _1112730034.unknown _1112730041.unknown _1112729932.unknown _1112729989.unknown _1112729900.unknown _1112729574.unknown _1112729680.unknown _1112729705.unknown _1112729307.unknown _1112729540.unknown _1112729144.unknown _1112729306.unknown _1112723446.unknown …
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1662926825.doc x b ab ab cd a a b d c ab ab ab ab ab ab cd cd ab c d a b ad ab cd ab cd a b c ab bc ac a b b c a c l ab 0 ¹ l ab l l ab l l ab ab l l ab l ab l ab o x ) ( x m o o o o m ом m x î í ì - = ом ом х x m ) ( x m 7 5 = - х x x x x ) ( 2 - n ) ( 12 m ) ( ), ( 2 2 1 1 x m x m 2 …

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