Thực hiện các phép tính sau:
a)x+1/x-3 – 1-2x/3+x – x(1-x)/9-x^2
b)3x+2/x^2-2x+1- 6/x^2-1- 3x-2/x^2+2x+1
c)1/a.(a-b).(a-c)+ 1/b.(b-a).(b-c)+ 1/c.(c-a).(c-b)
Thực hiện các phép tính sau:
a)x+1/x-3 – 1-2x/3+x – x(1-x)/9-x^2
b)3x+2/x^2-2x+1- 6/x^2-1- 3x-2/x^2+2x+1
c)1/a.(a-b).(a-c)+ 1/b.(b-a).(b-c)+ 1/c.(c-a).(c-b)
Đáp án:
$\begin{array}{l}
a)\dfrac{{x + 1}}{{x – 3}} – \dfrac{{1 – 2x}}{{3 + x}} – \dfrac{{x\left( {1 – x} \right)}}{{9 – {x^2}}}\\
= \dfrac{{\left( {x + 1} \right)\left( {3 + x} \right) – \left( {1 – 2x} \right)\left( {x – 3} \right) + x\left( {1 – x} \right)}}{{\left( {x – 3} \right)\left( {3 + x} \right)}}\\
= \dfrac{{{x^2} + 4x + 3 + 2{x^2} – 7x + 3 + x – {x^2}}}{{\left( {x – 3} \right)\left( {3 + x} \right)}}\\
= \dfrac{{2{x^2} – 2x + 6}}{{{x^2} – 9}}\\
b)\dfrac{{3x + 2}}{{{x^2} – 2x + 1}} – \dfrac{6}{{{x^2} – 1}} – \dfrac{{3x – 2}}{{{x^2} + 2x + 1}}\\
= \dfrac{{3x + 2}}{{{{\left( {x – 1} \right)}^2}}} – \dfrac{6}{{\left( {x – 1} \right)\left( {x + 1} \right)}} – \dfrac{{3x – 2}}{{{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{{\left( {3x + 2} \right){{\left( {x + 1} \right)}^2} – 6\left( {x + 1} \right)\left( {x – 1} \right) – \left( {3x – 2} \right){{\left( {x – 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}{{\left( {x – 1} \right)}^2}}}\\
= \dfrac{{\left( {3x + 2} \right)\left( {{x^2} + 2x + 1} \right) – 6{x^2} + 6 – \left( {3x – 2} \right)\left( {{x^2} – 2x + 1} \right)}}{{{{\left( {{x^2} – 1} \right)}^2}}}\\
= \dfrac{{3{x^3} + 8{x^2} + 7x + 2 – 6{x^2} + 6 – \left( {3{x^3} – 8{x^2} + 7x – 2} \right)}}{{{{\left( {{x^2} – 1} \right)}^2}}}\\
= \dfrac{{10{x^2} + 10}}{{{{\left( {{x^2} – 1} \right)}^2}}}\\
c)\dfrac{1}{{a\left( {a – b} \right)\left( {a – c} \right)}} + \dfrac{1}{{b\left( {b – a} \right)\left( {b – c} \right)}}\\
+ \dfrac{1}{{c\left( {c – a} \right)\left( {c – b} \right)}}\\
= \dfrac{1}{{a\left( {a – b} \right)\left( {a – c} \right)}} – \dfrac{1}{{b\left( {a – b} \right)\left( {b – c} \right)}}\\
+ \dfrac{1}{{c\left( {a – c} \right)\left( {b – c} \right)}}\\
= \dfrac{{b.c.\left( {b – c} \right) – a.c\left( {a – c} \right) + a.b\left( {a – b} \right)}}{{abc\left( {a – b} \right)\left( {b – c} \right).\left( {a – c} \right)}}\\
= \dfrac{{{b^2}c – b{c^2} – {a^2}c + a{c^2} + {a^2}b – a{b^2}}}{{abc\left( {a – b} \right)\left( {b – c} \right).\left( {a – c} \right)}}\\
= \dfrac{{{a^2}\left( {b – c} \right) + bc\left( {b – c} \right) – a\left( {{b^2} – {c^2}} \right)}}{{abc\left( {a – b} \right)\left( {b – c} \right).\left( {a – c} \right)}}\\
= \dfrac{{\left( {b – c} \right)\left( {{a^2} + bc – a\left( {b + c} \right)} \right)}}{{abc\left( {a – b} \right)\left( {b – c} \right).\left( {a – c} \right)}}\\
= \dfrac{{{a^2} + bc – ab – ac}}{{abc\left( {a – b} \right)\left( {a – c} \right)}}\\
= \dfrac{{a\left( {a – b} \right) – c\left( {a – b} \right)}}{{abc\left( {a – b} \right)\left( {a – c} \right)}}\\
= \dfrac{{\left( {a – b} \right)\left( {a – c} \right)}}{{abc\left( {a – b} \right)\left( {a – c} \right)}}\\
= \dfrac{1}{{abc}}
\end{array}$