3. Bài 3: Phép nhân, phép chia phân thức đại số

\(\begin{array}{l}A = \left( {1 – \frac{1}{{{2^2}}}} \right)\left( {1 – \frac{1}{{{3^2}}}} \right)\left( {1 – \frac{1}{{{4^2}}}} \right)\left( {1 – \frac{1}{{{5^2}}}} \right) \cdot  \cdot  \cdot \left( {1 – \frac{1}{{{n^2}}}} \right)\\ = \frac{{{2^2} – 1}}{{{2^2}}} \cdot \frac{{{3^2} – 1}}{{{3^2}}} \cdot \frac{{{4^2} – 1}}{{{4^2}}} \cdot \frac{{{5^2} – 1}}{{{5^2}}} \cdot  \cdot  \cdot \frac{{{n^2} – 1}}{{{n^2}}}\\ = \frac{{1.3}}{{{2^2}}} \cdot \frac{{2.4}}{{{3^2}}} \cdot \frac{{3.5}}{{{4^2}}} \cdot \frac{{4.6}}{{{5^2}}} \cdot  \cdot  \cdot \frac{{\left( {n – 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\\ = \frac{{1.2.3.4…\left( {n – 1} \right)}}{{2.3.4.5…n}} \cdot \frac{{3.4.5.6…\left( {n + 1} \right)}}{{2.3.4.5…n}}\\ = \frac{1}{n} \cdot \frac{{n + 1}}{2} = \frac{{n + 1}}{{2n}}\end{array}\)

Điều kiện: \(\left\{ \begin{array}{l}{x^2} – 1 \ne 0\\x + 4 \ne 0\\{x^2} + 6x \ne 0\\x – 4 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {x – 1} \right)\left( {x + 1} \right) \ne 0\\x + 4 \ne 0\\x\left( {x + 6} \right) \ne 0\\x – 4 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne  \pm 1\\x \ne  \pm 4\\x \ne 0\\x \ne  – 6\end{array} \right.\)

\(\begin{array}{l}\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}} = 0\\\frac{{x + 3}}{{{x^2} – 1}} \cdot \frac{{{x^2} + 6x}}{{x + 4}} – \frac{{x + 3}}{{{x^2} – 1}} \cdot \frac{{x – 4}}{{x + 4}} = 0\\\frac{{x + 3}}{{{x^2} – 1}}\left( {\frac{{{x^2} + 6x}}{{x + 4}} – \frac{{x – 4}}{{x + 4}}} \right) = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{\left( {{x^2} + 6x} \right) – \left( {x – 4} \right)}}{{x + 4}} = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{{x^2} + 6x – x + 4}}{{x + 4}} = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{{x^2} + 5x + 4}}{{x + 4}} = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{{x^2} + 4x + x + 4}}{{x + 4}} = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{x\left( {x + 4} \right) + \left( {x + 4} \right)}}{{x + 4}} = 0\\\frac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} \cdot \frac{{\left( {x + 1} \right)\left( {x + 4} \right)}}{{\left( {x + 4} \right)}} = 0\\\frac{{x + 3}}{{x – 1}} = 0\\x + 3 = 0\\x =  – 3\,\left( {{\rm{t/m}}} \right)\end{array}\)

Vậy có 1 giá trị của \(x\) thỏa mãn \(\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}} = 0\).

\(\begin{array}{l}A = \left( {4{x^2} – 16} \right) \cdot \frac{{7x – 2}}{{3x + 6}} = \frac{{\left( {4{x^2} – 16} \right)\left( {7x – 2} \right)}}{{3x + 6}} = \frac{{4\left( {x – 2} \right)\left( {x + 2} \right)\left( {7x – 2} \right)}}{{3\left( {x + 2} \right)}}\\ = \frac{{4\left( {x – 2} \right)\left( {7x – 2} \right)}}{3} = \frac{4}{3}\left( {7{x^2} – 2x – 14x + 4} \right) = \frac{4}{3}\left( {7{x^2} – 16x + 4} \right)\\ = \frac{4}{3}\left[ {{{\left( {\sqrt 7 x} \right)}^2} – 2 \cdot \sqrt 7 x \cdot \frac{8}{{\sqrt 7 }} + {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2} + 4 – {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2}} \right]\\ = \frac{4}{3}\left[ {{{\left( {\sqrt 7 x – \frac{8}{{\sqrt 7 }}} \right)}^2} – \frac{{36}}{7}} \right]\end{array}\)

Ta có: \({\left( {\sqrt 7 x – \frac{8}{{\sqrt 7 }}} \right)^2} \ge 0\forall x \Rightarrow {\left( {\sqrt 7 x – \frac{8}{{\sqrt 7 }}} \right)^2} – \frac{{36}}{7} \ge  – \frac{{36}}{7}\forall x\)

\(\frac{4}{3}\left[ {{{\left( {\sqrt 7 x – \frac{8}{{\sqrt 7 }}} \right)}^2} – \frac{{36}}{7}} \right] \ge \frac{4}{3} \cdot \left( { – \frac{{36}}{7}} \right) =  – \frac{{48}}{7}\) hay \(A \ge  – \frac{{48}}{7}\)

Dấu “=” xảy ra \( \Leftrightarrow {\left( {\sqrt 7 x – \frac{8}{{\sqrt 7 }}} \right)^2} = 0 \Leftrightarrow x = \frac{8}{7}\).

Vậy giá trị nhỏ nhất của biểu thức \(A = \left( {4{x^2} – 16} \right) \cdot \frac{{7x – 2}}{{3x + 6}}\) là \( – \frac{{48}}{7}\) khi \(x = \frac{8}{7}\).

\(\begin{array}{l}\left( {1 – \frac{1}{{{2^2}}}} \right)\left( {1 – \frac{1}{{{3^2}}}} \right)\left( {1 – \frac{1}{{{4^2}}}} \right)\left( {1 – \frac{1}{{{5^2}}}} \right) \cdot  \cdot  \cdot \left( {1 – \frac{1}{{{n^2}}}} \right) = \frac{{{2^2} – 1}}{{{2^2}}} \cdot \frac{{{3^2} – 1}}{{{3^2}}} \cdot \frac{{{4^2} – 1}}{{{4^2}}} \cdot \frac{{{5^2} – 1}}{{{5^2}}} \cdot  \cdot  \cdot \frac{{{n^2} – 1}}{{{n^2}}}\\ = \frac{{1.3}}{{{2^2}}} \cdot \frac{{2.4}}{{{3^2}}} \cdot \frac{{3.5}}{{{4^2}}} \cdot \frac{{4.6}}{{{5^2}}} \cdot  \cdot  \cdot \frac{{\left( {n – 1} \right)\left( {n + 1} \right)}}{{{n^2}}} = \frac{{1.2.3.4…\left( {n – 1} \right)}}{{2.3.4.5…n}} \cdot \frac{{3.4.5.6…\left( {n + 1} \right)}}{{2.3.4.5…n}} = \frac{1}{n} \cdot \frac{{n + 1}}{2} = \frac{{n + 1}}{{2n}}\end{array}\)Áp dụng với \(n = 2010\) ta có:

\(A = \left( {1 – \frac{1}{{{2^2}}}} \right)\left( {1 – \frac{1}{{{3^2}}}} \right) \cdot  \cdot  \cdot \left( {1 – \frac{1}{{{{2010}^2}}}} \right) = \frac{{2010 + 1}}{{2.2010}} = \frac{{2011}}{{4020}}\)

Ta có: \(1 – \frac{2}{{n\left( {n + 1} \right)}} = \frac{{{n^2} + n – 2}}{{n\left( {n + 1} \right)}} = \frac{{{n^2} + 2n – n – 2}}{{n\left( {n + 1} \right)}} = \frac{{n\left( {n + 2} \right) – \left( {n + 2} \right)}}{{n\left( {n + 1} \right)}} = \frac{{\left( {n – 1} \right)\left( {n + 2} \right)}}{{n\left( {n + 1} \right)}}\)\(\begin{array}{l}\left( {1 – \frac{2}{6}} \right)\left( {1 – \frac{2}{{12}}} \right) \cdot  \cdot  \cdot \left[ {1 – \frac{2}{{n\left( {n + 1} \right)}}} \right] = \frac{{1.4}}{{2.3}} \cdot \frac{{2.5}}{{3.4}} \cdot \frac{{3.6}}{{4.5}} \cdot  \cdot  \cdot \frac{{\left( {n – 1} \right)\left( {n + 2} \right)}}{{n\left( {n + 1} \right)}}\\ = \frac{{1.2.3…\left( {n – 1} \right)}}{{2.3.4…n}} \cdot \frac{{4.5.6…\left( {n + 2} \right)}}{{3.4.5…\left( {n + 1} \right)}} = \frac{1}{n} \cdot \frac{{n + 2}}{3} = \frac{{n + 2}}{{3n}}\\ = \frac{1}{3}\left( {\frac{{n + 2}}{n}} \right) = \frac{1}{3}\left( {1 + \frac{2}{n}} \right) > \frac{1}{3}\left( {1 + 0} \right) = \frac{1}{3}\left( {0 < \frac{2}{n} \le 1\forall n \ge 2} \right)\end{array}\)