A=[(2+x)/(2-x)-(4x^2)/(x^2-4)-(2-x)/(2+x)] ×[(6-x-x^2)/4(x^2-2x+1] RÚT GỌN a với giá trị nào của x thì a đạt giá trị nhỏ nhất

Giải thích các bước giải:

\(\begin{array}{l}
\left\{ \begin{array}{l}
x \ne  \pm 2\\
x \ne 1
\end{array} \right.\\
A = \left( {\frac{{2 + x}}{{2 – x}} – \frac{{4{x^2}}}{{{x^2} – 4}} – \frac{{2 – x}}{{2 + x}}} \right).\left( {\frac{{6 – x – {x^2}}}{{4\left( {{x^2} – 2x + 1} \right)}}} \right)\\
 = \left( {\frac{{{{\left( {2 + x} \right)}^2}}}{{\left( {2 – x} \right)\left( {2 + x} \right)}} – \frac{{4{x^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} – \frac{{{{\left( {2 – x} \right)}^2}}}{{\left( {2 + x} \right)\left( {2 – x} \right)}}} \right).\frac{{\left( {2 – x} \right)\left( {x + 3} \right)}}{{4{{\left( {x – 1} \right)}^2}}}\\
 = \frac{{{{\left( {2 + x} \right)}^2} + 4{x^2} – {{\left( {2 – x} \right)}^2}}}{{\left( {2 – x} \right)\left( {2 + x} \right)}}.\frac{{\left( {2 – x} \right)\left( {x + 3} \right)}}{{4{{\left( {x – 1} \right)}^2}}}\\
 = \frac{{{x^2} + 4x + 4 + 4{x^2} – 4 + 4x – {x^2}}}{{\left( {2 – x} \right)\left( {2 + x} \right)}}.\frac{{\left( {2 – x} \right)\left( {x + 3} \right)}}{{4{{\left( {x – 1} \right)}^2}}}\\
 = \frac{{4{x^2} + 8x}}{{\left( {2 + x} \right)}}.\frac{{x + 3}}{{4{{\left( {x – 1} \right)}^2}}}\\
 = \frac{{4x\left( {x + 2} \right)\left( {x + 3} \right)}}{{\left( {x + 2} \right).4.{{\left( {x – 1} \right)}^2}}}\\
 = \frac{{x\left( {x + 3} \right)}}{{{{\left( {x – 1} \right)}^2}}}
\end{array}\)