43В. Решите неравенство \(\frac{{{x^2} — 2x + 1}}{{{{\left( {x + 2} \right)}^2}}} + \frac{{{x^2} + 2x + 1}}{{{{\left( {x — 3} \right)}^2}}} \le \frac{{{{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}}\)
ОТВЕТ: \(\frac{1}{7}.\)
\(\frac{{{x^2} — 2x + 1}}{{{{\left( {x + 2} \right)}^2}}} + \frac{{{x^2} + 2x + 1}}{{{{\left( {x — 3} \right)}^2}}} \le \frac{{{{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\frac{{{{\left( {x — 1} \right)}^2}{{\left( {x — 3} \right)}^2}}}{{{{\left( {x — 3} \right)}^2}{{\left( {x + 2} \right)}^2}}} + \frac{{{{\left( {x + 1} \right)}^2}{{\left( {x + 2} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le \frac{{{{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\frac{{2{{\left( {{x^2} — 4x + 3} \right)}^2} + 2{{\left( {{x^2} + 3x + 2} \right)}^2} — {{\left( {2{x^2} — x + 5} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le 0.\) 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: \(\frac{{{{\left( {7x — 1} \right)}^2}}}{{2{{\left( {x + 2} \right)}^2}{{\left( {x — 3} \right)}^2}}} \le 0\;\;\;\; \Leftrightarrow \;\;\;\;x = \frac{1}{7}.\) Ответ: \(\frac{1}{7}.\)