Advertisements (Quảng cáo)
Theo định nghĩa của phép trừ, khi viết
\({A \over B} – {C \over D} – {E \over F}\) có nghĩa là \({A \over B} + {{ – C} \over D} + {{ – E} \over F}\)
Áp dụng điều này để làm các phép tính sau :
a. \({1 \over {3x – 2}} – {1 \over {3x + 2}} – {{3x – 6} \over {4 – 9{x^2}}}\)
b. \({{18} \over {\left( {x – 3} \right)\left( {{x^2} – 9} \right)}} – {3 \over {{x^2} – 6x + 9}} – {x \over {{x^2} – 9}}\)
Advertisements (Quảng cáo)
a. \({1 \over {3x – 2}} – {1 \over {3x + 2}} – {{3x – 6} \over {4 – 9{x^2}}}\)\( = {1 \over {3x – 2}} – {1 \over {3x + 2}} + {{3x – 6} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}}\)
\(\eqalign{ & = {{3x + 2} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}} + {{ – \left( {3x – 2} \right)} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}} + {{3x – 6} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}} \cr & = {{3x + 2 – 3x + 2 + 3x – 6} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}} = {{3x – 2} \over {\left( {3x + 2} \right)\left( {3x – 2} \right)}} = {1 \over {3x + 2}} \cr} \)
b. \({{18} \over {\left( {x – 3} \right)\left( {{x^2} – 9} \right)}} – {3 \over {{x^2} – 6x + 9}} – {x \over {{x^2} – 9}}\)\( = {{18} \over {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}} + {{ – 3} \over {{{\left( {x – 3} \right)}^2}}} + {{ – x} \over {\left( {x + 3} \right)\left( {x – 3} \right)}}\)
\(\eqalign{ & = {{18} \over {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}} + {{ – 3\left( {x + 3} \right)} \over {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}} + {{ – x\left( {x – 3} \right)} \over {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}} = {{18 – 3x – 9 – {x^2} + 3x} \over {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}} \cr & = {{9 – {x^2}} \over {\left( {3 – {x^2}} \right)\left( {x + 3} \right)}} = {{\left( {3 – x} \right)\left( {3 + x} \right)} \over {\left( {3 – {x^2}} \right)\left( {x + 3} \right)}} = {1 \over {3 – x}} \cr} \)