Bài 51 trang 58 SGK Toán 8 tập 1

Làm các phép tính sau:

a) \(\left( {\dfrac{{{x^2}}}{{{y^2}}} + \dfrac{y}{x}} \right):\left( {\dfrac{x}{{{y^2}}} - \dfrac{1}{y} + \dfrac{1}{x}} \right);\)

b) \(\left( {\dfrac{1}{{{x^2} + 4x + 4}} - \dfrac{1}{{{x^2} - 4x + 4}}} \right):\left( {\dfrac{1}{{x + 2}} + \dfrac{1}{{x - 2}}} \right)\)

\(\eqalign{
& a)\,\,\,\left( {{{{x^2}} \over {{y^2}}} + {y \over x}} \right):\left( {{x \over {{y^2}}} - {1 \over y} + {1 \over x}} \right) \cr
& = \left( {{{{x^2}.x} \over {x{y^2}}} + {{y.{y^2}} \over {x{y^2}}}} \right):\left( {{{x.x} \over {x{y^2}}} - {{xy} \over {x{y^2}}} + {{{y^2}} \over {x{y^2}}}} \right) \cr
& = {{{x^2}.x + y.{y^2}} \over {x{y^2}}}:{{{x^2} - xy + {y^2}} \over {x{y^2}}} \cr
& = {{{x^3} + {y^3}} \over {x{y^2}}}:{{{x^2} - xy + {y^2}} \over {x{y^2}}} \cr
& = {{{x^3} + {y^3}} \over {x{y^2}}}.{{x{y^2}} \over {{x^2} - xy + {y^2}}} \cr
& = {{\left( {{x^3} + {y^3}} \right)x{y^2}} \over {x{y^2}\left( {{x^2} - xy + {y^2}} \right)}} \cr
& = {{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)x{y^2}} \over {x{y^2}\left( {{x^2} - xy + {y^2}} \right)}} \cr
& = x + y \cr} \)

\(\eqalign{
& b)\;\left( {{1 \over {{x^2} + 4x + 4}} - {1 \over {{x^2} - 4x + 4}}} \right):\left( {{1 \over {x + 2}} + {1 \over {x - 2}}} \right) \cr
& = \left( {{1 \over {{x^2} + 2.x.2 + {2^2}}} - {1 \over {{x^2} - 2.x.2 + {2^2}}}} \right):\left( {{1 \over {x + 2}} + {1 \over {x - 2}}} \right) \cr
& = \left[ {{1 \over {{{\left( {x + 2} \right)}^2}}} - {1 \over {{{\left( {x - 2} \right)}^2}}}} \right]:{{x - 2 + x + 2} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} \cr
& = {{{{\left( {x - 2} \right)}^2} - {{\left( {x + 2} \right)}^2}} \over {{{\left( {x + 2} \right)}^2}{{\left( {x - 2} \right)}^2}}}.{{\left( {x + 2} \right)\left( {x - 2} \right)} \over {2x}} \cr
& = {{\left[ {{x^2} - 4x + 4 - \left( {{x^2} + 4x + 4} \right)} \right]\left( {x + 2} \right)\left( {x - 2} \right)} \over {2x{{(x + 2)}^2}{{(x - 2)}^2}}} \cr
& = {{\left( {{x^2} - 4x + 4 - {x^2} - 4x - 4} \right)\left( {x + 2} \right)\left( {x - 2} \right)} \over {2x{{(x + 2)}^2}{{(x - 2)}^2}}} \cr
& = {{\left( { - 8x} \right)\left( {x + 2} \right)\left( {x - 2} \right)} \over {2x{{(x + 2)}^2}{{(x - 2)}^2}}} \cr
& = {{ - 4} \over {(x + 2)(x - 2)}} = {{ - 4} \over {{x^2} - 4}} \cr} \)