Question

ayuda, de panas doy corona. Es mi califacion del periodo

Answer 1

Respuesta:

Suma de polinomios

a)

7a - 4b + 5c - 7a + 4b - 6c

-c

b)

9x - 3y + 5 - x - y + 4 - 5x + 4y - 9

3x

c)

5x - 7y + 8 - y + 6 - 4x + 9 - 3x + 8y

-2x + 23

d)

3x^2 - 4xy + y^2 - 5xy + 6x^2 - 3y^2 - 6y^2 - 8xy - 9

$9x^2 - 17xy - 8y^2 - 9$

ordenas

$9x^2 - 8y^2 - 17xy - 9$

e)

x^3 + xy + y^3 - 5x^2y + x^3 - y^3 + 2x^3 - 4xy^2 - 5y^3

$4x^3 + xy - 5y^3 - 5x^2y - 4xy^2$

ordenas

$4x^3 - 5x^2y - 4xy^2 - 5y^3 + xy$

f)

$x^5 - x^3y^2 - xy^4 +2x^4y + 3x^2y^3 - y^5 + 3x^3y^2 - 4xy^4 - y^5 + x^5 +5xy^4 + 2y^5$

$2x^5 + 2x^3y^2 + 2x^4y + 3x^2y^3$

ordenas

$2x^5 + 2x^4y + 2x^2y^3+ 3x^3y^2$

g)

$x^2 +\frac{2}{3} xy -\frac{1}{6}xy +y^2-\frac{5}{6}xy +\frac{2}{3}y^2\\\\ x^2 + \frac{4 - 1 - 5}{6}xy + \frac{3 + 2}{3}y^2 \\\\x^2 -\frac{2}{6}xy + \frac{5}{3}y^2\\\\x^2 + \frac{5}{3}y^2 - \frac{1}{3}xy$

h)

$-\frac{3}{4}x^2 -\frac{1}{2}y^2 -\frac{2}{5}xy +\frac{1}{6}y^2 +\frac{1}{10}xy +\frac{1}{3}y^2\\\\-\frac{3}{4}x^2 +\frac{-3 + 1 + 2}{6}y^2+ \frac{-4 +1}{10}xy\\\\-\frac{3}{4}x^2 + \frac{0}{6}y^2 + \frac{-3}{10}xy\\\\-\frac{3}{4}x^2 - \frac{3}{10}xy$

i)

$x^3 - 9x + 6x^2 - 19-(-11x^2+21x -43 + 6x^3)\\\\x^3 - 9x + 6x^2 - 19 + 11x^2 - 21x + 43 - 6x^3\\\\-5x^3- 30x + 17x^2 + 24\\\\-5x^3 + 17x^2 - 30x + 24$

j)

$-a^5b+6a^3b^3-18ab^5+4 - (-8a^6+9b^6-11a^4b^2-11a^2b^4)\\\\-a^5b + 6a^3b^3 - 18ab^5 + 4 + 8a^6 - 9ab^6 + 11a^4b^2 + 11a^2b^2\\\\8a^6-a^5b+11a^4b^2+ 6a^3b^3+11a^2b^4-18ab^5-9b^6+4$

Explicación paso a paso:

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