Câu 26 trang 31 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn biểu thức :

a. \({{3{x^2} + 5x + 1} \over {{x^3} - 1}} - {{1 - x} \over {{x^2} + x + 1}} - {3 \over {x - 1}}\)

b. \({1 \over {{x^2} - x + 1}} + 1 - {{{x^2} + 2} \over {{x^3} + 1}}\)

c. \({7 \over x} - {x \over {x + 6}} + {{36} \over {{x^2} + 6x}}\)

Giải:

a. \({{3{x^2} + 5x + 1} \over {{x^3} - 1}} - {{1 - x} \over {{x^2} + x + 1}} - {3 \over {x - 1}}\)

\(\eqalign{  &  = {{3{x^2} + 5x + 1} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{x - 1} \over {{x^2} + x + 1}} + {{ - 3} \over {x - 1}}  \cr  &  = {{3{x^2} + 5x + 1} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{{{\left( {x - 1} \right)}^2}} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ - 3\left( {{x^2} + x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}  \cr  &  = {{3{x^2} + 5x + 1 + {x^2} - 2x + 1 - 3{x^2} - 3x - 3} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{{x^2} - 1} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}  \cr  &  = {{\left( {x + 1} \right)\left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{x + 1} \over {{x^2} + x + 1}} \cr} \)

b. \({1 \over {{x^2} - x + 1}} + 1 - {{{x^2} + 2} \over {{x^3} + 1}}\)\( = {1 \over {{x^2} - x + 1}} + 1 + {{ - \left( {{x^2} + 2} \right)} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)

\(\eqalign{  &  = {{x + 1} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + {{{x^3} + 1} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + {{ - \left( {{x^2} + 2} \right)} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}  \cr  &  = {{x + 1 + {x^3} + 1 - {x^2} - 2} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = {{x + {x^3} - {x^2}} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = {{x\left( {{x^2} - x + 1} \right)} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = {x \over {x + 1}} \cr} \)

c. \({7 \over x} - {x \over {x + 6}} + {{36} \over {{x^2} + 6x}}\)\( = {7 \over x} + {{ - x} \over {x + 6}} + {{36} \over {{x^2} + 6x}} = {{7\left( {x + 6} \right)} \over {x\left( {x + 6} \right)}} + {{ - {x^2}} \over {x\left( {x + 6} \right)}} + {{36} \over {x\left( {x + 6} \right)}}\)

\(\eqalign{  &  = {{7x + 42 - {x^2} + 36} \over {x\left( {x + 6} \right)}} = {{7x - {x^2} + 78} \over {x\left( {x + 6} \right)}} = {{13x + 78 - 6x - {x^2}} \over {x\left( {x + 6} \right)}}  \cr  &  = {{13\left( {x + 6} \right) - x\left( {x + 6} \right)} \over {x\left( {x + 6} \right)}} = {{\left( {x + 6} \right)\left( {13 - x} \right)} \over {x\left( {x + 6} \right)}} = {{13 - x} \over x} \cr} \)

Sachbaitap.com