Advertisements (Quảng cáo)
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. \({7 \over x} – {x \over {x + 6}} + {{36} \over {{x^2} + 6x}}\)
a. \({{3{x^2} + 5x + 1} \over {{x^3} – 1}} – {{1 – x} \over {{x^2} + x + 1}} – {3 \over {x – 1}}\)
\(\eqalign{ & = {{3{x^2} + 5x + 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{x – 1} \over {{x^2} + x + 1}} + {{ – 3} \over {x – 1}} \cr & = {{3{x^2} + 5x + 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{{{\left( {x – 1} \right)}^2}} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ – 3\left( {{x^2} + x + 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & = {{3{x^2} + 5x + 1 + {x^2} – 2x + 1 – 3{x^2} – 3x – 3} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{{x^2} – 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & = {{\left( {x + 1} \right)\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{x + 1} \over {{x^2} + x + 1}} \cr} \)
Advertisements (Quảng cáo)
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. \({7 \over x} – {x \over {x + 6}} + {{36} \over {{x^2} + 6x}}\)\( = {7 \over x} + {{ – x} \over {x + 6}} + {{36} \over {{x^2} + 6x}} = {{7\left( {x + 6} \right)} \over {x\left( {x + 6} \right)}} + {{ – {x^2}} \over {x\left( {x + 6} \right)}} + {{36} \over {x\left( {x + 6} \right)}}\)
\(\eqalign{ & = {{7x + 42 – {x^2} + 36} \over {x\left( {x + 6} \right)}} = {{7x – {x^2} + 78} \over {x\left( {x + 6} \right)}} = {{13x + 78 – 6x – {x^2}} \over {x\left( {x + 6} \right)}} \cr & = {{13\left( {x + 6} \right) – x\left( {x + 6} \right)} \over {x\left( {x + 6} \right)}} = {{\left( {x + 6} \right)\left( {13 – x} \right)} \over {x\left( {x + 6} \right)}} = {{13 – x} \over x} \cr} \)