Thực hiện phép tính:a) 4/x+2+3/x-2+5x+2/4-x² b) 1+x/x-3+1-2x/3+x+x(1-x)/9-x² c) 3x+2/x²-2x+1+6/x²-1+3x-2/x²+2x+1

Đáp án:

a) \(\dfrac{2}{{x + 2}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{4}{{x + 2}} + \dfrac{3}{{x – 2}} + \dfrac{{5x + 2}}{{4 – {x^2}}}\\
 = \dfrac{{4\left( {x – 2} \right) + 3\left( {x + 2} \right) – 5x – 2}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{4x – 8 + 3x + 6 – 5x – 2}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{2x – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{2}{{x + 2}}\\
b)\dfrac{{1 + x}}{{x – 3}} + \dfrac{{1 – 2x}}{{3 + x}} + \dfrac{{x\left( {1 – x} \right)}}{{9 – {x^2}}}\\
 = \dfrac{{\left( {x + 1} \right)\left( {x – 1} \right) + \left( {1 – 2x} \right)\left( {x – 3} \right) – x + {x^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{x^2} – 1 – 2{x^2} + 7x – 3 – x + {x^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{6x – 4}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
c)\dfrac{{3x + 2}}{{{x^2} – 2x + 1}} + \dfrac{6}{{{x^2} – 1}} + \dfrac{{3x – 2}}{{{x^2} + 2x + 1}}\\
 = \dfrac{{\left( {3x + 2} \right)\left( {{x^2} + 2x + 1} \right) + 6\left( {{x^2} – 1} \right) + \left( {3x – 2} \right)\left( {{x^2} – 2x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x – 1} \right)}^2}}}\\
 = \dfrac{{3{x^3} + 6{x^2} + 3x + 2{x^2} + 4x + 2 + 6{x^2} – 6 + 3{x^3} – 6{x^2} + 3x – 2{x^2} + 4x – 2}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x – 1} \right)}^2}}}\\
 = \dfrac{{6{x^3} + 6{x^2} + 14x – 6}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x – 1} \right)}^2}}}
\end{array}\)