Bài 1. Trừ các phân thức:
a) \({{2x} \over {x – 4}} – {{5x – 2} \over {{x^2} – 16}}\)
b) \({{2x + 8} \over {{x^2} – 4x + 4}} – {7 \over {x – 2}}\)
c) \(x – {{xy} \over {x + y}} – {{{x^3}} \over {{x^2} – {y^2}}}.\)
Bài 2. Chứng minh rằng: \({{3{a^2} + 3} \over {{a^3} – 1}} – {{a – 1} \over {{a^2} + a + 1}} + {2 \over {1 – a}} = 0.\)
Bài 1. a) \(MTC = {x^2} – 16 = \left( {x – 4} \right)\left( {x + 4} \right).\)
Vậy \({{2x} \over {x – 4}} – {{5x – 2} \over {{x^2} – 16}} = {{2x\left( {x + 4} \right) – \left( {5x – 2} \right)} \over {{x^2} – 16}}\)
Advertisements (Quảng cáo)
\( = {{2{x^2} + 8x – 5x + 2} \over {{x^2} – 16}} = {{2{x^2} + 3x + 2} \over {{x^2} – 16}}.\)
b) \(MTC = {x^2} – 4x + 4 = {\left( {x – 2} \right)^2}.\)
Vậy \({{2x + 8} \over {{x^2} – 4x + 4}} – {7 \over {x – 2}} = {{2x + 8 – 7\left( {x – 2} \right)} \over {{{\left( {x – 2} \right)}^2}}}\)
\( = {{2x + 8 – 7x + 14} \over {{{\left( {x – 2} \right)}^2}}} = {{22 – 5x} \over {{{\left( {x – 2} \right)}^2}}}.\)
c) \(MTC = {x^2} – 4x + 4 = {\left( {x – 2} \right)^2}.\)
Advertisements (Quảng cáo)
Vậy \({{2x + 8} \over {{x^2} – 4x + 4}} – {7 \over {x – 2}} = {{2x + 8 – 7\left( {x – 2} \right)} \over {{{\left( {x – 2} \right)}^2}}}\)
\( = {{2x + 8 – 7x + 14} \over {{{\left( {x – 2} \right)}^2}}} = {{22 – 5x} \over {{{\left( {x – 2} \right)}^2}}}\) .
c) \(MTC = {x^2} – {y^2} = \left( {x – y} \right)\left( {x + y} \right).\)
Vậy \(x – {{xy} \over {x + y}} – {{{x^3}} \over {{x^2} – {y^2}}} = {{x\left( {{x^2} – {y^2}} \right) – xy\left( {x – y} \right) – {x^3}} \over {{x^2} – {y^2}}}\)
\( = {{{x^2} – x{y^2} – {x^2}y + x{y^2} – {x^3}} \over {{x^2} – {y^2}}} = {{ – {x^2}y} \over {{x^2} – {y^2}}} = – {{{x^2}y} \over {{x^2} – {y^2}}}.\)
Bài 2. Biến đổi vế trái (VT), ta được:
\(MTC = {a^3} – 1 = \left( {a – 1} \right)\left( {{a^2} + a + 1} \right)\)
\(VT = {{3{a^2} + 3 – {{\left( {a – 1} \right)}^2} – 2\left( {{a^2} + a + 1} \right)} \over {{a^3} – 1}}\)
\( = {{3{a^2} + 3 – {a^2} + 2a – 1 – 2{a^2} – 2a – 2} \over {{a^3} – 1}} = 0 = VP\).
