Question

Надо! сократить дробь: а) 39x³y ¯¯¯¯¯¯¯ = 26x²y² б) 5y ¯¯¯¯¯¯ = y²—2y в) 3a—3b ¯¯¯¯¯¯¯¯¯¯ = a²—b² представить в виде дроби: а) 3a—2a 1—a² ¯¯¯¯¯¯¯¯¯¯ — ¯¯¯¯¯¯¯¯ = 2a a² б) 1 1 ¯¯¯¯¯¯¯¯ — ¯¯¯¯¯¯ = 3x+y 3x—y в) 4—3b 3 ¯¯¯¯¯¯¯¯¯¯ + ¯¯¯¯¯¯ = b²—2b b—2 найти значения выражений: а) при x= —8, y=0,1 x—6y² ¯¯¯¯¯¯¯¯ + 3y = 2y

Answer 1

$\frac{39x^3y}{26x^2y^2}= \frac{13*3*x^2*x*y}{13*2*x^2y*y}= \frac{3*x}{2*y}= \frac{3x}{2y}$

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$\frac{5y}{y^2-2y} = \frac{5*y}{y*y-y*2} = \frac{5*y}{y*(y-2)} = \frac{5}{y-2}$

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

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

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$\frac{1}{3x+y}- \frac{1}{3x-y} = \frac{1*(3x-y)}{(3x+y)(3x-y)}- \frac{1*(3x+y)}{(3x-y)(3x+y)} =$
$= \frac{(3x-y)-(3x+y)}{(3x+y)(3x-y)} = \frac{3x-y-3x-y}{(3x)^2-(y)^2} = \frac{-y-y}{3^2x^2-y^2} = \frac{-2y}{9x^2-y^2} = -\frac{2y}{9x^2-y^2}$

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$\frac{4-3b}{b^2-2b}+ \frac{3}{b-2}= \frac{4-3b}{b*b-b*2}+ \frac{3}{b-2}= \frac{4-3b}{b*(b-2)}+ \frac{3*b}{(b-2)*b}= \frac{4-3b+3b}{b*(b-2)}= \frac{4}{b(b-2)}$

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$\frac{x-6y^2}{2y}+3y= \frac{x-6y^2}{2y}+3y* \frac{2y}{2y} = \frac{x-6y^2}{2y}+\frac{3y*2y}{2y} = \frac{x-6y^2}{2y}+\frac{6y^2}{2y} =$
$= \frac{x-6y^2+6y^2}{2y}= \frac{2x}{y} =(2x):y=(2*(-8)):0.1=-16: \frac{1}{10}=$
$=-16* \frac{10}{1}= -16*10=-160$