Trắc nghiệm Bài 23: Phép cộng và phép trừ phân thức đại số Toán 8 Kết nối tri thức

\(\begin{array}{l}A = \frac{3}{{2{x^2} + 2x}} + \frac{{\left| {2x - 1} \right|}}{{{x^2} - 1}} - \frac{2}{x} = \frac{3}{{2x\left( {x + 1} \right)}} + \frac{{2x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{2}{x}\\ = \frac{{3\left( {x - 1} \right) + 2x\left( {2x - 1} \right) - 4\left( {x - 1} \right)\left( {x + 1} \right)}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}}\\ = \frac{{3x - 3 + 4{x^2} - 2x - 4{x^2} + 4}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{x + 1}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2x\left( {x - 1} \right)}}\end{array}\)

\(\begin{array}{l}\frac{1}{{1 - x}} + \frac{1}{{1 + x}} + \frac{2}{{1 + {x^2}}} + \frac{4}{{1 + {x^4}}} + \frac{8}{{1 + {x^8}}} = \frac{{1 + x + 1 - x}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} + \frac{2}{{1 + {x^2}}} + \frac{4}{{1 + {x^4}}} + \frac{8}{{1 + {x^8}}}\\ = \frac{2}{{1 - {x^2}}} + \frac{2}{{1 + {x^2}}} + \frac{4}{{1 + {x^4}}} + \frac{8}{{1 + {x^8}}} = \frac{{2\left( {1 + {x^2}} \right) + 2\left( {1 - {x^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 + {x^2}} \right)}} + \frac{4}{{1 + {x^4}}} + \frac{8}{{1 + {x^8}}}\\ = \frac{4}{{1 - {x^4}}} + \frac{4}{{1 + {x^4}}} + \frac{8}{{1 + {x^8}}} = \frac{{4\left( {1 + {x^4}} \right) + 4\left( {1 - {x^4}} \right)}}{{\left( {1 - {x^4}} \right)\left( {1 + {x^4}} \right)}} + \frac{8}{{1 + {x^8}}}\\ = \frac{8}{{1 - {x^8}}} + \frac{8}{{1 + {x^8}}} = \frac{{8\left( {1 + {x^8}} \right) + 8\left( {1 - {x^8}} \right)}}{{\left( {1 - {x^8}} \right)\left( {1 + {x^8}} \right)}} = \frac{{16}}{{1 - {x^{16}}}}\end{array}\)

Thay \(2023 = abc\) vào biểu thức \(A\) ta được:

\(\begin{array}{l}\frac{{2023a}}{{ab + 2023a + 2023}} + \frac{b}{{bc + b + 2023}} + \frac{c}{{ac + 1 + c}}\\ = \frac{{{a^2}bc}}{{ab + {a^2}bc + abc}} + \frac{b}{{bc + b + abc}} + \frac{c}{{ac + 1 + c}}\\ = \frac{{{a^2}bc}}{{ab\left( {1 + ac + c} \right)}} + \frac{b}{{b\left( {c + 1 + ac} \right)}} + \frac{c}{{ac + 1 + c}}\\ = \frac{{ac}}{{1 + ac + c}} + \frac{1}{{c + 1 + ac}} + \frac{c}{{ac + 1 + c}} = 1\end{array}\)

\(\begin{array}{l}\frac{{{a^2}}}{{{{\left( {b - c} \right)}^2}}} + \frac{{{b^2}}}{{{{\left( {c - a} \right)}^2}}} + \frac{{{c^2}}}{{{{\left( {a - b} \right)}^2}}} = {\left( {\frac{a}{{b - c}}} \right)^2} + {\left( {\frac{b}{{c - a}}} \right)^2} + {\left( {\frac{c}{{a - b}}} \right)^2}\\ = {\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{a - b}}} \right)^2} - 2\left[ {\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ca}}{{\left( {a - b} \right)\left( {b - c} \right)}}} \right]\\ \ge  - 2\left[ {\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ca}}{{\left( {a - b} \right)\left( {b - c} \right)}}} \right]\end{array}\)

(Vì \({\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{a - b}}} \right)^2} \ge 0\forall a,\,b,\,c\) đôi một khác nhau)

Mà \(\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ac}}{{\left( {a - b} \right)\left( {b - c} \right)}}\)

\(\begin{array}{l} = \frac{{ab\left( {a - b} \right) + bc\left( {b - c} \right) + ac\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\ = \frac{{ab\left( {a - b} \right) + bc\left( {b - c} \right) + ac\left( {c - b + b - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\ = \frac{{\left( {ab - ac} \right)\left( {a - b} \right) + \left( {bc - ac} \right)\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\ = \frac{{a\left( {b - c} \right)\left( {a - b} \right) - c\left( {a - b} \right)\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\ = \frac{{\left( {a - c} \right)\left( {a - b} \right)\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} =  - 1\end{array}\)

\(\begin{array}{l} \Rightarrow \frac{{{a^2}}}{{{{\left( {b - c} \right)}^2}}} + \frac{{{b^2}}}{{{{\left( {c - a} \right)}^2}}} + \frac{{{c^2}}}{{{{\left( {a - b} \right)}^2}}}\\ \ge  - 2\left[ {\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ca}}{{\left( {a - b} \right)\left( {b - c} \right)}}} \right]\\ = \left( { - 2} \right)\left( { - 1} \right) = 2\end{array}\)