пож помогите 266 (2) и 268 (1). Created by goldsvetlana140. algebra-ru

266(2)1/[a(a-c)-b(a-c)]+2/[b(b-a)-c(b-a)]+1/[c(c-a)-b(c-a)]==1/(a-c)(a-b)-2/(a-b)(b-c)+1/(a-c)(b-c)=(b-c-2a+2c+a-b)/(a-b)(a-c)(b-c)==(c-a)/(a-b)(a-c)(b-c)=1/(b-a)(b-c)268(1)1/x(x-1)(x-2)=a/x+b/(x-1)+c(x+2)1=a(x-1)(x+2)+bx(x+2)+cx(x-1)1=ax²+ax-2a+bx²+2bx+cx²-cx1=x²(a+b+c)+x(a+2b-c)-2a{a+b+c=0{a+2b-c=0{-2a=1⇒a=-1/2прибавим 1 и 22a+3b=03b=-2a3b=-2*(-1/2)=1b=1/3подставим в 1-1/2+1/3+с=0с=1/2-1/3=3/6-2/6=1/6Ответ a=-1/2;b=1/3;c=1/6Новые вопросы в АлгебраАлгебра ДПА 9 класс 2 вариант 1.7Алгебра Мерзляк ДПА 9 класс 1.9 1 вариантАлгебра Мерзляк ДПА 9 классДопоможітьбудь ласка Знайдіть суму 100 перших членів послідовності аn, якщо ат = 2n+1Варіанти відповідей1020020400120001020a1=-5; d=6. Скільки треба взяти членів прогресії, щоб їх сума дорівнювала 35?Варіанти відповідей25730Помогите пожайлуста даю 70 балов!!!Верно ли,что если сумма a² + b²(a и b натуральные)делятся на 11,то числа a и b делятся на 11?Срочно с решениемПЛИСПлизХ - 38=378СпасайтеПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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пож помогите 266 (2) и 268 (1)...

goldsvetlana140 @goldsvetlana140

July 2022 1 11 Report

пож помогите 266 (2) и 268 (1)

Answers & Comments

sedinalana 266(2)
1/[a(a-c)-b(a-c)]+2/[b(b-a)-c(b-a)]+1/[c(c-a)-b(c-a)]=
=1/(a-c)(a-b)-2/(a-b)(b-c)+1/(a-c)(b-c)=(b-c-2a+2c+a-b)/(a-b)(a-c)(b-c)=
=(c-a)/(a-b)(a-c)(b-c)=1/(b-a)(b-c)
268(1)
1/x(x-1)(x-2)=a/x+b/(x-1)+c(x+2)
1=a(x-1)(x+2)+bx(x+2)+cx(x-1)
1=ax²+ax-2a+bx²+2bx+cx²-cx
1=x²(a+b+c)+x(a+2b-c)-2a
{a+b+c=0
{a+2b-c=0
{-2a=1⇒a=-1/2
прибавим 1 и 2
2a+3b=0
3b=-2a
3b=-2*(-1/2)=1
b=1/3
подставим в 1
-1/2+1/3+с=0
с=1/2-1/3=3/6-2/6=1/6
Ответ a=-1/2;b=1/3;c=1/6