\(\dfrac{{{x^2} — 2x + 1}}{{{{\left( {x + 2} \right)}^2}}} + \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x — 3} \right)}^2}}} \le \dfrac{{{{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}}\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\dfrac{{{{\left( {x — 1} \right)}^2}{{\left( {x — 3} \right)}^2}}}{{{{\left( {x — 3} \right)}^2}{{\left( {x + 2} \right)}^2}}} + \dfrac{{{{\left( {x + 1} \right)}^2}{{\left( {x + 2} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le \dfrac{{{{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}}\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\dfrac{{2{{\left( {{x^2} — 4x + 3} \right)}^2} + 2{{\left( {{x^2} + 3x + 2} \right)}^2} — {{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le 0.\)
Преобразуем числитель. Пусть \({x^2} — 4x + 3 = a,\;\;\;\;{x^2} + 3x + 2 = b.\) Тогда:
\(a + b = {x^2} — 4x + 3 + {x^2} + 3x + 2 = 2{x^2} — x + 5.\)
\(2{\left( {{x^2} — 4x + 3} \right)^2} + 2{\left( {{x^2} + 3x + 2} \right)^2} — {\left( {2{x^2} — x + 5} \right)^2} = 2{a^2} + 2{b^2} — {\left( {a + b} \right)^2} = \)
\( = 2{a^2} + 2{b^2} — {a^2} — 2ab — {b^2} = {a^2} — 2ab + {b^2} = {\left( {a — b} \right)^2} = \)
\( = {\left( {{x^2} — 4x + 3 — {x^2} — 3x — 2} \right)^2} = {\left( {7x — 1} \right)^2}.\)
Тогда исходное неравенство примет вид:
\(\dfrac{{{{\left( {7x — 1} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le 0\;\;\;\; \Leftrightarrow \;\;\;\;x = \dfrac{1}{7}.\)
Ответ: \(\dfrac{1}{7}.\)