Здравствуйте, 10 класс алгебра, 3 номер, если хотите можете решить и 4, но можно только большое.

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Ответ:

Kira2236

05.10.2020

1)-2x+1=-x-6
-2x+x=-6-1
-x=-7
x=7

2)3/8x=24;
x=24:3/8
x=24×8/3
x=64

2(0,6x+1,85)=1,3x+0,7
1,2x+3,7=1,3x+0,7
1,2x-1,3x=0,7-3,7
-0,1x=-3
x=30

Найдите значение числового выражения:
(2/7+3/14)(7,5-13,5)=

1) 2/7×7,5-2/7×13,5+3/14×7,5-3/14×13,5=

2) 7,5=7 5/10=7 1/2=(1+7×2)/2=15/2

3)2/7×15/2=15/2

4)13,5=13 1/2=(1+13×2)/2=27/2

5)2/7×27/2=27/7

6)3/14×7,5=3/14×15/2=45/28

7)3/14×27/2=81/28
8)15/2-27/7+45/28-81/28=
105/14-54/14-36/28=51/14-36/28=102/28-36/28=66/28=33/14=2 5/14

Упростите выражение:
1)5a-3b-8a+12b
-3a+9b

2)16c+(3c-2)-(5c+7)
16c+3c-2-5c-7=14c-9

3)7-3(6y-4)
7-18y+12=-18y+19

4,8(91 оценок)

Ответ:

Dima25688

05.10.2020

-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)

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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)

4,7(75 оценок)

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