Bài 58 trang 62 SGK Toán 8 tập 1

Thực hiện các phép tính sau:

\(\eqalign{
& a)\,\,\left( {{{2x + 1} \over {2x - 1}} - {{2x - 1} \over {2x + 1}}} \right):{{4x} \over {10x - 5}} \cr
& b)\,\,\left( {{1 \over {{x^2} + x}} - {{2 - x} \over {x + 1}}} \right):\left( {{1 \over x} + x - 2} \right) \cr
& c){\rm{ }}{1 \over {x - 1}} - {{{x^3} - x} \over {{x^2} + 1}}.\left( {{1 \over {{x^2} - 2x + 1}} + {1 \over {1 - {x^2}}}} \right). \cr} \)

\(\eqalign{
& a)\,\,\left( {{{2x + 1} \over {2x - 1}} - {{2x - 1} \over {2x + 1}}} \right):{{4x} \over {10x - 5}} \cr
& = {{{{\left( {2x + 1} \right)}^2} - {{\left( {2x - 1} \right)}^2}} \over {\left( {2x - 1} \right)\left( {2x + 1} \right)}}.{{10x - 5} \over {4x}} \cr
& = {{4{x^2} + 4x + 1 - 4{x^2} + 4x - 1} \over {\left( {2x - 1} \right)\left( {2x + 1} \right)}}.{{5\left( {2x - 1} \right)} \over {4x}} \cr
& = {{8x.5\left( {2x - 1} \right)} \over {\left( {2x - 1} \right)\left( {2x + 1} \right).4x}} = {{10} \over {2x + 1}} \cr} \)

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

\(\eqalign{
& c)\;\,{1 \over {x - 1}} - {{{x^3} - x} \over {{x^2} + 1}}.\left( {{1 \over {{x^2} - 2x + 1}} + {1 \over {1 - {x^2}}}} \right) \cr
&={1 \over {x - 1}} - {{{x^3} - x} \over {{x^2} + 1}}.\left[ {{1 \over {{{\left( {x - 1} \right)}^2}}} - {1 \over {{x^2} - 1}}} \right]\cr&= {1 \over {x - 1}} - {{{x^3} - x} \over {{x^2} + 1}}.\left[ {{1 \over {{{\left( {x - 1} \right)}^2}}} - {1 \over {\left( {x - 1} \right)\left( {x + 1} \right)}}} \right] \cr
& = {1 \over {x - 1}} - {{x\left( {{x^2} - 1} \right)} \over {{x^2} + 1}}.{{x + 1 - \left( {x - 1} \right)} \over {{{\left( {x - 1} \right)}^2}.\left( {x + 1} \right)}} \cr
& = {1 \over {x - 1}} - {{x\left( {x - 1} \right)\left( {x + 1} \right)} \over {{x^2} + 1}}.{{x + 1 - x + 1} \over {{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} \cr
& = {1 \over {x - 1}} - {{x\left( {x - 1} \right)\left( {x + 1} \right).2} \over {\left( {{x^2} + 1} \right){{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} \cr
& = {1 \over {x - 1}} - {{2x} \over {\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} \cr
& = {{{x^2} + 1 - 2x} \over {\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} \cr
& = {{{{\left( {x - 1} \right)}^2}} \over {\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} \cr
& = {{x - 1} \over {{x^2} + 1}} \cr} \)