Алгебра: Доказать, что функция f(x) непрерывна в точке f(x)= =3

Доказать, что функция f(x) непрерывна в точке f(x)= =3

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Ответ:

potato1999

07.09.2020

$lim _{x-\ \textgreater \ 3+0} = -5*(3+0)^2-9 = -54 \\
 lim _{x-\ \textgreater \ 3-0} = -5(3-0)^2-9 = -54 \\
$
$limit_{x-\ \textgreater \ 3} \ -5*(3)^2-9=-54$ предел совпадает со значением в точке
то есть предел в точке

существует , значит функция непрерывна в этой точке

4,5(26 оценок)

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Ответ:

Ксюника1

07.09.2020

а) (а – 2)( а + 2) – 2а(5 – а) =а^2-4-10a+2a^2=6a^2-10a-4
б) (у – 9)2 – 3у(у + 1) =y^2-18y+81-3y^2-3y=-2y^2-21y+81
в) 3(х – 4) 2 – 3х2 =3(x^2-8x+16)-3x^2=3x^2-24x+48-3x^2=48-24x
2. Разложите на множители:
а) 25х – х3=x(25-x^2)=x(5-x)(5+x) б) 2х2 – 20х + 50 =2(x^2-10x+25)=2(x-5)^2=2(x-5)(x+5)
3. Найдите значение выражения а2 – 4bс=36-4*(-11)*(-10)=36-440=-404
а) 452 б) -202 в) -404 г) 476
4. Упростите выражение:
(с2 – b)2 – (с2 - 1)(с2 + 1) + 2bс2 =c^4-4bc^2+b^2-c^4+1=-4bc^2+b^2+1
5. Докажите тождество:
(а + b)2 – (а – b)2 = 4аba^2+2ab+b^2-a^2+2ab-b^2=2a+2ab=4ab

4,5(80 оценок)

Ответ:

Dima25688

07.09.2020

-a4b+a4c+a3b2+2a3bc-2a3c2-2a2b3-a2b2c-a2bc2+a2c3+ab4+2ab3c-ab2c2+2abc3-ac4-b4c+b3c2-2b2c3+bc4 ————————————————————————————————————————————————————————————————————————————————————————————— (a-b)•(b-c)•(c-a) Reformatting the input :

Changes made to your input should not affect the solution:

(1): "c2" was replaced by "c^2". 2 more similar replacement(s).

Step by step solution :Skip Ad
Step 1 : c2 Simplify ————— c - a Equation at the end of step 1 : (a2) (b2) c2 ((—————•(a-c))+(—————•(b-a)))+(———•(c-b)) (a-b) (b-c) c-a Step 2 :Equation at the end of step 2 : (a2) (b2) c2•(c-b) ((—————•(a-c))+(—————•(b-a)))+———————— (a-b) (b-c) c-a Step 3 : b2 Simplify ————— b - c Equation at the end of step 3 : (a2) b2 c2•(c-b) ((—————•(a-c))+(———•(b-a)))+———————— (a-b) b-c c-a Step 4 :Equation at the end of step 4 : (a2) b2•(b-a) c2•(c-b) ((—————•(a-c))+————————)+———————— (a-b) b-c c-a Step 5 : a2 Simplify ————— a - b Equation at the end of step 5 : a2 b2•(b-a) c2•(c-b) ((———•(a-c))+————————)+———————— a-b b-c c-a Step 6 :Equation at the end of step 6 : a2•(a-c) b2•(b-a) c2•(c-b) (————————+————————)+———————— a-b b-c c-a Step 7 :Calculating the Least Common Multiple :

7.1    Find the Least Common Multiple

      The left denominator is :       a-b

      The right denominator is :       b-c

                  Number of times each Algebraic Factor
            appears in the factorization of:    Algebraic
    Factor     Left
Denominator  Right
Denominator  L.C.M = Max
{Left,Right}  a-b 101 b-c 011

Least Common Multiple:
(a-b) • (b-c)

Calculating Multipliers :

7.2    Calculate multipliers for the two fractions

    Denote the Least Common Multiple by  L.C.M
    Denote the Left Multiplier by  Left_M
    Denote the Right Multiplier by  Right_M
    Denote the Left Deniminator by  L_Deno
    Denote the Right Multiplier by  R_Deno

   Left_M = L.C.M / L_Deno = b-c

   Right_M = L.C.M / R_Deno = a-b

Making Equivalent Fractions :

7.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. a2 • (a-c) • (b-c) —————————————————— = —————————————————— L.C.M (a-b) • (b-c) R. Mult. • R. Num. b2 • (b-a) • (a-b) —————————————————— = —————————————————— L.C.M (a-b) • (b-c) Adding fractions that have a common denominator :

7.4 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

a2 • (a-c) • (b-c) + b2 • (b-a) • (a-b) a3b - a3c - a2b2 - a2bc + a2c2 + 2ab3 - b4 ——————————————————————————————————————— = —————————————————————————————————————————— (a-b) • (b-c) (a - b) • (b - c) Equation at the end of step 7 : (a3b - a3c - a2b2 - a2bc + a2c2 + 2ab3 - b4) c2 • (c - b) ———————————————————————————————————————————— + ———————————— (a - b) • (b - c) c - a Step 8 :Calculating the Least Common Multiple :

8.1    Find the Least Common Multiple

      The left denominator is :       (a-b) • (b-c)

      The right denominator is :       c-a

                  Number of times each Algebraic Factor
            appears in the factorization of:    Algebraic
    Factor     Left
Denominator  Right
Denominator  L.C.M = Max
{Left,Right}  a-b 101 b-c 101 c-a 011

Least Common Multiple:
(a-b) • (b-c) • (c-a)

Calculating Multipliers :

8.2    Calculate multipliers for the two fractions

    Denote the Least Common Multiple by  L.C.M
    Denote the Left Multiplier by  Left_M
    Denote the Right Multiplier by  Right_M
    Denote the Left Deniminator by  L_Deno
    Denote the Right Multiplier by  R_Deno

   Left_M = L.C.M / L_Deno = c-a

   Right_M = L.C.M / R_Deno = (a-b)•(b-c)

Making Equivalent Fractions :

8.3 Rewrite the two fractions into equivalent fractions

L. Mult. • L. Num. (a3b-a3c-a2b2-a2bc+a2c2+2ab3-b4) • (c-a) —————————————————— = ———————————————————————————————————————— L.C.M (a-b) • (b-c) • (c-a) R. Mult. • R. Num. c2 • (c-b) • (a-b) • (b-c) —————————————————— = —————————————————————————— L.C.M (a-b) • (b-c) • (c-a) Adding fractions that have a common denominator :

8.4 Adding up the two equivalent fractions

(a3b-a3c-a2b2-a2bc+a2c2+2ab3-b4) • (c-a) + c2 • (c-b) • (a-b) • (b-c) -a4b+a4c+a3b2+2a3bc-2a3c2-2a2b3-a2b2c-a2bc2+a2c3+ab4+2ab3c-ab2c2+2abc3-ac4-b4c+b3c2-2b2c3+bc4 ————————————————————————————————————————————————————————————————————— = ————————————————————————————————————————————————————————————————————————————————————————————— (a-b) • (b-c) • (c-a) (a-b) • (b-c) • (c-a) Final result : -a4b+a4c+a3b2+2a3bc-2a3c2-2a2b3-a2b2c-a2bc2+a2c3+ab4+2ab3c-ab2c2+2abc3-ac4-b4c+b3c2-2b2c3+bc4 ————————————————————————————————————————————————————————————————————————————————————————————— (a-b)•(b-c)•(c-a)

Processing ends successfully

Latest drills solved(-4,7)to(94,-55)(5)/(7)+(4)/(y)=38(x+8/9)-9a2/(a-b)(a-c)+b2/(b-c)(b-a)+c2/(c-a)(c-b)

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