Question

qual o resultado da soma pela diferença
(x2y3/z + 5/6).(x2y3/z - 5/6)

Answer 1

Utilizando a mesma propriedade teremos:

$(\frac{ x^{2} y^{3} }{z}+ \frac{5}{6} )*( \frac{ x^{2} y^{3} }{z}- \frac{5}{6}) =( \frac{ x^{2} y^{3} }{z})² - (\frac{5}{6} )² = \frac{ x^{4} y^{6} }{ z^{2} } - \frac{25}{36}$

Se não for isso avise
 $\frac{12 a^{6} b^{9} c^{4} }{ -9a^{8} b^{7} c^{6} } = \frac{-4 b^{2} }{3a^{2} c^{2} }$

$\frac{14 a^{2} - 7ax}{10ay - 5xy} = \frac{7a(2a-x)}{5y(2a-x)}= \frac{7a}{5y}$

$\frac{12 a^{2} xy}{6ab-4 a^{2} c} = \frac{12 a^{2} xy}{2a(3b-2ac)}= \frac{6 a xy}{(3b-2ac)}$

$\frac{x^{2} -16 y^{2}}{8ay-2ax} = \frac{(x-4y)(x+4y)}{2a(4y-x)} = \frac{(x-4y)(x+4y)}{-2a(x-4y)} = \frac{(x+4y)}{-2a}= \frac{-x-4y}{2a}$

$\frac{ a^{2} -2a+1}{ a^{2} -1} = \frac{(a-1)(a-1)}{(a-1)(a+1)} = \frac{a-1}{a+1}$

$\frac{ a^{2} (x+y)}{a x^{2} -a y^{2} } = \frac{a^{2} (x+y)}{a(x^{2} - y^{2})} = \frac{a^{2} (x+y)}{a(x-y)(x+y)} = \frac{a}{(x-y)}$

$\frac{-2 a^{6} }{4 a^{2} b^{3}-2a b^{2} } = \frac{-2 a^{6}}{2a b^{2} (2ab-1)}= \frac{-a^{5} }{b^{2} (2ab-1)}$

$\frac{ a^{2} +ab}{ a^{2} -ab} = \frac{a(a+b)}{a(a-b)} = \frac{a+b}{a-b}$

$\frac{ x^{2} +2ax+ a^{2} }{mx+ma}= \frac{(x+a)(x+a)}{m(x+a)} = \frac{x+a}{m}$


$\frac{ a^{2} +a}{ a^{2} -9} + \frac{2}{a-3} + \frac{2}{a+3} = \frac{ a^{2} +a}{(a+3)(a-3)}+ \frac{2(a+3)}{(a+3)(a-3)} + \frac{2(a-3)}{(a+3)(a-3)}$$= \frac{ a^{2} +a+2a+6+2a-6}{(a+3)(a-3)}$$= \frac{a^{2}+5a}{ a^{2} -9}$

$\frac{3 x^{2} +3}{3x-3} - \frac{2x}{x-1} = \frac{3( x^{2} +1)}{3(x-1)} - \frac{2x}{x-1} = \frac{x^{2} +1}{x-1} - \frac{2x}{x-1}= \frac{ x^{2} -2x+1}{x-1}$$= \frac{ (x-1)(x-1)}{x-1} = x-1$

$\frac{a+b}{a-b} + \frac{a-2b}{a+b} = \frac{(a+b)(a+b)}{(a+b)(a-b)} + \frac{(a-b)(a-2b)}{(a+b)(a-b)} = = \frac{ a^{2} +2ab+ b^{2} +a^{2}-3ab+2 b^{2} }{(a+b)(a-b)}$$= \frac{ 2a^{2} -3ab+3 b^{2} }{ a^{2} - b^{2} }$

$\frac{x-y}{y} + \frac{x+y}{x} + \frac{ x^{2} -y^{2}}{2xy} = \frac{ 2 x^{2} -2xy}{2xy}+\frac{ 2xy+2 y^{2} }{2xy}+\frac{ x^{2} -y^{2}}{2xy}$$= \frac{ 2 x^{2} -2xy+2xy+2 y^{2}+x^{2} -y^{2}}{2xy}} = \frac{ 3x^{2} +y^{2}}{2xy}$

$\frac{2}{x+1} + \frac{3}{ x^{2} +x} + \frac{1}{x}= \frac{2x}{x(x+1)} + \frac{3}{ x(x+1)} + \frac{x+1}{x(x+1)} = \frac{3x+4}{ x^{2} +x}$

$\frac{3a}{x} - \frac{5a}{4x}+ \frac{7a}{2x} = \frac{4*3a}{4x}- \frac{5a}{4x}+ \frac{2*7a}{4x} = \frac{12a-5a+14a}{4x}= \frac{21a}{4x}$