Question

Помогите пожалуйста решить номер 37 (д,е)
Номер 38 (г)
Номер 40 (весь)
Номер 39 (б,е,а)
Кто все решит дам 50 баллов)

Answer 1

37
д)
$\frac{2x-1}{x^2-6x+9} : \frac{2-4x}{x^2-3x} = \frac{2x-1}{x^2-6x+9} * \frac{x^2-3x}{2-4x} =\frac{2x-1}{(x-3)^2} * \frac{x(x-3)}{2(1-2x)} = \\ \frac{-(1-2x)}{x-3} * \frac{x}{2(1-2x)} =\frac{-1}{x-3} * \frac{x}{2} =- \frac{x}{2(x-3)} =- \frac{x}{2x-6} =\frac{x}{6-2x}$

e)
$\frac{2a-4}{a^2+4} : \frac{a^2-4a+4}{a^4-16} = \frac{2(a-2)}{a^2+4} * \frac{a^4-16}{a^2-4a+4} =\frac{2(a-2)}{a^2+4} * \frac{(a^2)^2-4^2}{(a-2)^2} = \\ =\frac{2}{a^2+4} * \frac{(a^2-4)(a^2+4)}{a-2} =\frac{2}{1} * \frac{a^2-4}{a-2} =2*\frac{(a-2)(a+2)}{a-2}=2*\frac{a+2}{1}=2a+4$

38 г)
$(a-4b)^2:(32b^2-2a^2)= \frac{(a-4b)^2}{32b^2-2a^2} = \frac{(a-4b)^2}{2(16b^2-a^2)}= \frac{(4b-a)^2}{2((4b)^2-a^2)}= \\ \frac{(4b-a)^2}{2(4b-a)(4b+a)}= \frac{4b-a}{2(4b+a)}= \frac{4b-a}{8b+2a}$

39
a)
$\frac{a^2+ab+b^2}{x-2} : \frac{a^3-b^3}{x^2-4}= \frac{a^2+ab+b^2}{x-2}* \frac{x^2-4}{a^3-b^3} = \frac{a^2+ab+b^2}{x-2}* \frac{(x-2)(x+2)}{(a-b)(a^2+ab+b^2)} = \\ \frac{1}{1}* \frac{x+2}{a-b} = \frac{x+2}{a-b}$

б)
$\frac{ax^2-9a}{x^3+8} : \frac{x-3}{2x+4} = \frac{ax^2-9a}{x^3+8} * \frac{2x+4} {x-3} =\frac{a(x^2-9)}{x^3+2^3} * \frac{2(x+2)} {x-3} = \\ \frac{a(x-3)(x+3)}{(x+2)(x^2-2x+4)} *\frac{2(x+2)} {x-3} = \frac{a(x+3)}{x^2-2x+4} *\frac{2} {1} =\frac{2ax+6a}{x^2-2x+4}$

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

40
a)
$( \frac{x}{y^2}- \frac{1}{x} ):( \frac{1}{y}- \frac{1}{x} )=( \frac{x*x}{x*y^2}- \frac{y^2}{x*y^2} ):( \frac{x}{xy}- \frac{y}{xy} )= \\ \frac{x^2-y^2}{xy^2}: \frac{x-y}{xy} =\frac{(x-y)(x+y)}{xy^2}*\frac{xy} {x-y}=\frac{x+y}{y}*\frac{1} {1}=\frac{x+y}{y}$

б)
$( \frac{x}{x-1}-1)* \frac{xy-y}{x} =( \frac{x}{x-1}-\frac{x-1}{x-1})* \frac{y(x-1)}{x} =\frac{x-x+1}{x-1}* \frac{y(x-1)}{x} =\frac{1}{1}* \frac{y}{x} = \\ =y/x$

в)
$\frac{a}{b} - \frac{a^2-b^2}{b^2}: \frac{a+b}{b}= \frac{a}{b} - \frac{(a-b)(a+b)}{b^2}* \frac{b}{a+b}= \frac{a}{b} - \frac{a-b}{b}* \frac{1}{1}= \frac{a}{b} - \frac{a-b}{b}= \\ \frac{a-a+b}{b}=b/b=1$

г)
$(\frac{x+3}{x-3}- \frac{x-3}{x+3}): \frac{12x}{x^2+6x+9} = (\frac{(x+3)^2}{(x+3)(x-3)}- \frac{(x-3)^2}{(x+3)(x-3)}): \frac{12x}{(x+3)^2} = \\ \frac{(x+3)^2-(x-3)^2}{(x+3)(x-3)}* \frac{(x+3)^2}{12x} =\frac{(x+3)^2-(x-3)^2}{x-3}* \frac{x+3}{12x} = \\ \frac{((x+3)-(x-3))((x+3)+(x-3))}{x-3}* \frac{x+3}{12x} =\frac{(x+3-x+3)(x+3+x-3-)}{x-3}* \frac{x+3}{12x} = \\ \frac{6*2x}{x-3}* \frac{x+3}{12x} =\frac{12x}{x-3}* \frac{x+3}{12x} =\frac{1}{x-3}* \frac{x+3}{1} =\frac{x+3}{x-3}$

фух... :-)