1) thực hiên phép tính : a) 1/x-1 – x^3-x/ x^2+1 nhân ( 1/x^2-2x+1 + 1/1-x^2 ) b) ( 2x-3/x(x+1)^2 + 4-x/x(x+1)^2 ) : 4/3x^2+3x

Đáp án:

b) \(\dfrac{3}{4}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{1}{{x – 1}} – \dfrac{{x\left( {{x^2} – 1} \right)}}{{{x^2} + 1}}.\left( {\dfrac{1}{{{x^2} – 2x + 1}} + \dfrac{1}{{1 – {x^2}}}} \right)\\
 = \dfrac{1}{{x – 1}} – \dfrac{{x\left( {{x^2} – 1} \right)}}{{{x^2} + 1}}.\left[ {\dfrac{{x + 1 – x + 1}}{{\left( {{x^2} – 1} \right)\left( {x – 1} \right)}}} \right]\\
 = \dfrac{1}{{x – 1}} – \dfrac{{x\left( {{x^2} – 1} \right)}}{{{x^2} + 1}}.\dfrac{2}{{\left( {{x^2} – 1} \right)\left( {x – 1} \right)}}\\
 = \dfrac{1}{{x – 1}} – \dfrac{{2x}}{{\left( {x – 1} \right)\left( {{x^2} + 1} \right)}}\\
 = \dfrac{{{x^2} + 1 – 2x}}{{\left( {x – 1} \right)\left( {{x^2} + 1} \right)}}\\
 = \dfrac{{{{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {{x^2} + 1} \right)}} = \dfrac{{x – 1}}{{{x^2} + 1}}\\
b)\left( {\dfrac{{2x – 3}}{{x{{\left( {x + 1} \right)}^2}}} + \dfrac{{4 – x}}{{x{{\left( {x + 1} \right)}^2}}}} \right):\dfrac{4}{{3x\left( {x + 1} \right)}}\\
 = \dfrac{{2x – 3 + 4 – x}}{{x{{\left( {x + 1} \right)}^2}}}.\dfrac{{3x\left( {x + 1} \right)}}{4}\\
 = \dfrac{{x + 1}}{{x{{\left( {x + 1} \right)}^2}}}.\dfrac{{3x\left( {x + 1} \right)}}{4}\\
 = \dfrac{3}{4}
\end{array}\)