Ответ:
[tex]\dfrac{ 3x ^ { 2 } }{ x ^ { 2 } -1 } - \dfrac{ 3x }{ x-1 } =\dfrac{3x^{2} }{(x-1)(x+1)} - \dfrac{ 3x }{ x-1 }=\dfrac{3x^2-3x(x+1)}{(x-1)(x+1)} =\dfrac{3x^2-3x^2-3x}{(x-1)(x+1)}=\dfrac{-3x}{x^{2}-1}[/tex]
[tex]\dfrac{ 2y ^ { 2 } }{ y-8 } -2y = \dfrac{ 2y ^ { 2 } }{ y-8 } -\dfrac{2y( y-8)}{ y-8} = \dfrac{ 2y ^ { 2 } -2y( y-8) }{ y-8 } =\dfrac{2y^2-2y^2+16y}{y-8}= \dfrac{16y}{y-8}[/tex]
[tex]\dfrac{ 5a+5b }{ { b }^{ 2 } } *\dfrac{ b }{ a+b } = \dfrac{ 5(a+b) }{ { b }^{ 2 } } *\dfrac{ b }{ a+b } =\dfrac{5}{b}[/tex]
[tex]( \dfrac{ -2a ^ { 3 } }{ b ^ { 4 } } ) ^ { 2 } =\dfrac{4*a^6}{b^8} =\dfrac{4a^6}{b^8}[/tex]
[tex]\dfrac{ y }{ xy-x } : \dfrac{ y }{ y ^ { 2 } -1 } =\dfrac{y(y ^ { 2 } -1 )}{y(xy-x )} =\dfrac{y^2-1}{x(y-1)} =\dfrac{(y-1)(y+1)}{x(y-1)}=\dfrac{y+1}{x}[/tex]
[tex]\dfrac{ (a ^ { 2 } -x ^ { 2 } ) }{ 2a+ \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2}{a} + \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2}{a} + \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2+2x }{a} }=\dfrac{a(a ^ { 2 } -x ^ { 2 } )}{2(a^2+x)}[/tex]
Объяснение:
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Ответ:
[tex]\dfrac{ 3x ^ { 2 } }{ x ^ { 2 } -1 } - \dfrac{ 3x }{ x-1 } =\dfrac{3x^{2} }{(x-1)(x+1)} - \dfrac{ 3x }{ x-1 }=\dfrac{3x^2-3x(x+1)}{(x-1)(x+1)} =\dfrac{3x^2-3x^2-3x}{(x-1)(x+1)}=\dfrac{-3x}{x^{2}-1}[/tex]
[tex]\dfrac{ 2y ^ { 2 } }{ y-8 } -2y = \dfrac{ 2y ^ { 2 } }{ y-8 } -\dfrac{2y( y-8)}{ y-8} = \dfrac{ 2y ^ { 2 } -2y( y-8) }{ y-8 } =\dfrac{2y^2-2y^2+16y}{y-8}= \dfrac{16y}{y-8}[/tex]
[tex]\dfrac{ 5a+5b }{ { b }^{ 2 } } *\dfrac{ b }{ a+b } = \dfrac{ 5(a+b) }{ { b }^{ 2 } } *\dfrac{ b }{ a+b } =\dfrac{5}{b}[/tex]
[tex]( \dfrac{ -2a ^ { 3 } }{ b ^ { 4 } } ) ^ { 2 } =\dfrac{4*a^6}{b^8} =\dfrac{4a^6}{b^8}[/tex]
[tex]\dfrac{ y }{ xy-x } : \dfrac{ y }{ y ^ { 2 } -1 } =\dfrac{y(y ^ { 2 } -1 )}{y(xy-x )} =\dfrac{y^2-1}{x(y-1)} =\dfrac{(y-1)(y+1)}{x(y-1)}=\dfrac{y+1}{x}[/tex]
[tex]\dfrac{ (a ^ { 2 } -x ^ { 2 } ) }{ 2a+ \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2}{a} + \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2}{a} + \dfrac{ 2x }{ a } } =\dfrac{ (a ^ { 2 } -x ^ { 2 } )}{\dfrac{2a^2+2x }{a} }=\dfrac{a(a ^ { 2 } -x ^ { 2 } )}{2(a^2+x)}[/tex]
Объяснение: