УПРОСТИТЬ ВЫРАЖЕНИЯ. 1.[latex]( \frac{b}{9a-a^3}- \frac{1}{a^2+3a}+ \frac{3}{a^2b-9b}): \frac{b^2-6b+9}{a^3b-9ab} [/latex] 2.[latex](1- \frac{1}{1-a}):(1- \frac{1-2a^2}{1-a}+a)[/latex] 3.[latex] \frac{a^2+a-ab-b}{a^2+a+ab+b}: ...

[latex]( \frac{b}{9a-a^3}- \frac{1}{a^2+3a}+ \frac{3}{a^2b-9b}): \frac{b^2-6b+9}{a^3b-9ab} = \\\ =( \frac{b}{a(9-a^2)}- \frac{1}{a(a+3)}+ \frac{3}{b(a^2-9)}): \frac{(b-3)^2}{ab(a^2-9)} = \\\ =( -\frac{b}{a(a-3)(a+3)}- \frac{1}{a(a+3)}+ \frac{3}{b(a-3)(a+3)})\cdot \frac{ab(a^2-9)}{(b-3)^2} = \\\ =\frac{-b^2-ab+3b+3a}{ab(a-3)(a+3)}\cdot \frac{ab(a^2-9)}{(b-3)^2} = \frac{-b(b+a)+3(a+b)}{ab(a-3)(a+3)}\cdot \frac{ab(a^2-9)}{(b-3)^2} = \\\ =\frac{(a+b)(3-b)ab(a^2-9)}{ab(a-3)(a+3)(3-b)^2} = \frac{a+b}{3-b}[/latex] [latex](1- \frac{1}{1-a}):(1- \frac{1-2a^2}{1-a}+a)= \frac{1-a-1}{1-a}: \frac{1-a-1+2a^2+a-a^2}{1-a}= \\\ =- \frac{a}{1-a}: \frac{a^2}{1-a}=- \frac{a}{1-a}\cdot \frac{1-a}{a^2}=- \frac{a}{2a^2}=- \frac{1}{a}[/latex] [latex] \frac{a^2+a-ab-b}{a^2+a+ab+b}: \frac{a^2-a-ab+b}{a^2-a+ab-b} = \frac{a(a+1)-b(a+1)}{a(a+1)+b(a+1)}: \frac{a(a-1)-b(a-1)}{a(a-1)+b(a-1)} = \\\ = \frac{(a+1)(a-b)}{(a+1)(a+b)}: \frac{(a-1)(a-b)}{(a-1)(a+b)} = \frac{a-b}{a+b}: \frac{a-b}{a+b} = \frac{a-b}{a+b}\cdot \frac{a+b}{a-b} =1[/latex] [latex]( \frac{6}{a^2+5a+4}- \frac{2}{a^2+3a+2}+ \frac{a}{a^2+6a+8})^2\cdot \frac{a^2+4a+4}{2} = \\\ =( \frac{6}{a^2+4a+a+4}- \frac{2}{a^2+2a+a+2}+ \frac{a}{a^2+4a+2a+8})^2\cdot\frac{(a+2)^2}{2} = \\\ =( \frac{6}{a(a+4)+a+4}- \frac{2}{a(a+2)+a+2}+ \frac{a}{a(a+4)+2(a+4)})^2\cdot\frac{(a+2)^2}{2} = \\\ =( \frac{6}{(a+4)(a+1)}- \frac{2}{(a+2)(a+1)}+ \frac{a}{(a+4)(a+2)})^2\cdot\frac{(a+2)^2}{2} =[/latex] [latex]=( \frac{6a+12-2a-8+a^2+a}{(a+4)(a+1)(a+2)})^2\cdot\frac{(a+2)^2}{2} = ( \frac{a^2+4a+a+4}{(a+4)(a+1)(a+2)})^2\cdot\frac{(a+2)^2}{2} = \\\ =( \frac{a(a+4)+a+4}{(a+4)(a+1)(a+2)})^2\cdot\frac{(a+2)^2}{2} = ( \frac{(a+4)(a+1)}{(a+4)(a+1)(a+2)})^2\cdot\frac{(a+2)^2}{2} = \\\ =( \frac{1}{a+2})^2\cdot\frac{(a+2)^2}{2} = \frac{1}{(a+2)^2}\cdot\frac{(a+2)^2}{2} =\frac{1}{2}[/latex] [latex]\frac{(ab^-^1+ a^{-1}b+1)(a^-^1-b^-^1)^2 }{a^2b^-^2+a^-^2b^2-(ab^-^1+a^-^1b)} = \dfrac{( \frac{a}{b} + \frac{b}{a} +1)( \frac{1}{a} - \frac{1}{b} )^2 }{ \frac{a^2}{b^2} + \frac{b^2}{a^2} -( \frac{a}{b} + \frac{b}{a} )} \\\ = \dfrac{( \frac{a^2+b^2+ab}{ab} )( \frac{b-a}{ab})^2 }{ \frac{a^4+b^4}{a^2b^2} -( \frac{a^2+b^2}{ab} )} =\dfrac{\frac{a^2+b^2+ab}{ab}\cdot\frac{(b-a)^2}{(ab)^2} }{ \frac{a^4+b^4-ab(a^2+b^2)}{a^2b^2}} = \frac{(a^2+b^2+ab)(b-a)^2}{ab(a^4+b^4-a^3b-ab^3)} =[/latex] [latex]= \frac{(a^2+b^2+ab)(b-a)^2}{ab(a^3(a-b)-b^3(a-b))} = \frac{(a^2+b^2+ab)(a-b)^2}{ab(a-b)(a^3-b^3)} = \\\ = \frac{(a^2+b^2+ab)(a-b)^2}{ab(a-b)(a-b)(a^2+ab+b^2)} = \frac{1}{ab} [/latex]