Lời giải

Đáp án đúng là: D

\(\frac{{x + y}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\frac{{{x^2} + xy - 2{y^2}}}{{{x^4} - {y^4}}}\)

\( = \frac{{x + y}}{{{x^2}\left( {x + y} \right) + {y^2}\left( {x + y} \right)}}:\frac{{{x^2} + 2xy - xy - 2{y^2}}}{{\left( {{x^2} - {y^2}} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{{x + y}}{{\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}:\frac{{x\left( {x + 2y} \right) - y\left( {x + 2y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}}:\frac{{\left( {x - y} \right)\left( {x + 2y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}}:\frac{{x + 2y}}{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}} \cdot \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}{{x + 2y}} = \frac{{x + y}}{{x + 2y}}\).

Ta có \(\frac{{x + y}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\frac{{{x^2} + xy - 2{y^2}}}{{{x^4} - {y^4}}} = 2\)

\(\frac{{x + y}}{{x + 2y}} = 2\)

\[x + y = 2x + 4y\]

\[x = - 3y\]

Lời giải

Đáp án đúng là: B

\(\frac{{3x + 15}}{{{x^2} - 4}}:\frac{{x + 5}}{{x - 2}} = \frac{{3\left( {x + 5} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\frac{{x + 5}}{{x - 2}}\)

\( = \frac{{3\left( {x + 5} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \cdot \frac{{x - 2}}{{x + 5}} = \frac{3}{{x + 2}}\)

Khi đó \(\frac{{3x + 15}}{{{x^2} - 4}}:\frac{{x + 5}}{{x - 2}} = 1\)

\(\frac{3}{{x + 2}} = 1\)

\(x + 2 = 3\)

\(x = 1\) (TM)

Lời giải

Đáp án đúng là: D

\(A = \frac{{{x^2} + {y^2} + xy}}{{{x^2} - {y^2}}}:\frac{{{x^3} - {y^3}}}{{{x^2} + {y^2} - 2xy}}\)

\( = \frac{{{x^2} + {y^2} + xy}}{{\left( {x + y} \right)\left( {x - y} \right)}}:\frac{{\left( {x - y} \right)\left( {{x^2} + {y^2} + xy} \right)}}{{{{\left( {x - y} \right)}^2}}}\)

\( = \frac{{{x^2} + {y^2} + xy}}{{\left( {x + y} \right)\left( {x - y} \right)}} \cdot \frac{{{{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {{x^2} + {y^2} + xy} \right)}} = \frac{1}{{x + y}}\).

Với x + y = 5 ta có \[{\rm{A}} = \frac{1}{5}\]

\(B = \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}:\frac{{{x^2} - 2xy + {y^2}}}{{{x^4} - {y^4}}}\)

\( = \frac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + {y^2}}}:\frac{{{{\left( {x - y} \right)}^2}}}{{\left( {{x^2} + {y^2}} \right)\left( {x - y} \right)\left( {x + y} \right)}}\)

\( = \frac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + {y^2}}} \cdot \frac{{\left( {{x^2} + {y^2}} \right)\left( {x - y} \right)\left( {x + y} \right)}}{{{{\left( {x - y} \right)}^2}}} = {\left( {x + y} \right)^2}\)

Với x + y = 5 ta có \[{\rm{B}} = {5^2} = 25\]

Lời giải

\[{\rm{A}} = \frac{{{5^2} - 1}}{{{3^2} - 1}}:\frac{{{9^2} - 1}}{{{7^2} - 1}}:\frac{{{{13}^2} - 1}}{{{{11}^2} - 1}}:...:\frac{{{{55}^2} - 1}}{{{{53}^2} - 1}}\]

\[ = \frac{{{5^2} - 1}}{{{3^2} - 1}} \cdot \frac{{{7^2} - 1}}{{{9^2} - 1}} \cdot \frac{{{{11}^2} - 1}}{{{{13}^2} - 1}}...\frac{{{{53}^2} - 1}}{{{{55}^2} - 1}}\]

\[ = \frac{{4.6}}{{2.4}} \cdot \frac{{6.8}}{{8.10}} \cdot \frac{{10.12}}{{12.14}}...\frac{{52.54}}{{54.56}}\]

\[ = \frac{6}{2} \cdot \frac{6}{{10}} \cdot \frac{{10}}{{14}}...\frac{{52}}{{56}}\]

\[ = 3 \cdot \frac{6}{{56}} = \frac{9}{{28}}\]

Đáp án đúng là: A

Lời giải

Đáp án đúng là: A

\(A:B = \frac{{{x^3} - {x^2} - x + 11}}{{x - 2}}:\frac{{x + 2}}{{x - 2}} = \frac{{{x^3} - {x^2} - x + 11}}{{x - 2}} \cdot \frac{{x - 2}}{{x + 2}}\)

\( = \frac{{{x^3} - {x^2} - x + 11}}{{x + 2}} = \frac{{{x^3} + 2{x^2} - 3{x^2} - 6x + 5x + 10 + 1}}{{x + 2}}\)

\( = \frac{{{x^2}\left( {x + 2} \right) - 3x\left( {x + 2} \right) + 5\left( {x + 2} \right) + 1}}{{x + 2}}\)

\( = \frac{{\left( {x + 2} \right)\left( {{x^2} - 3x + 5} \right) + 1}}{{x + 2}} = {x^2} - 3x + 5 + \frac{1}{{x + 2}}\).

Để phân thức A chia hết cho phân thức B thì \[\frac{{\rm{A}}}{{\rm{B}}} \in \mathbb{Z}\].

Suy ra \(\left( {{x^2} - 3x + 5 + \frac{1}{{x + 2}}} \right) \in \mathbb{Z}\)

Mà \(\left( {{x^2} - 3x + 5} \right) \in \mathbb{Z}\,\,\,\forall x \in \mathbb{Z}\) hay \(\left( {x + 2} \right) \in U\left( 1 \right) = \left\{ { \pm \,1} \right\}\)

\(\left[ {\begin{array}{*{20}{c}}{x + 2 = - 1}\\{x + 2 = 1}\end{array}} \right.\)

\(\left[ {\begin{array}{*{20}{c}}{x = - 3}\\{x = - 1}\end{array}\,\,} \right.\,(TM)\)

Lời giải

Đáp án đúng là: A

Do \[a + b + c = 0\] nên \[a = - \left( {b + c} \right)\].

• \[4bc - {a^2} = 4bc - {\left[ { - \left( {b + c} \right)} \right]^2} = 4bc - \left( {{b^2} + 2bc + {c^2}} \right)\]

\[ = 2bc - {b^2} - {c^2} = - {\left( {b - c} \right)^2}\]

• \[bc + 2{a^2} = {a^2} + bc + {a^2} = {a^2} + bc + a\left[ { - \left( {b + c} \right)} \right]\]

\[ = {a^2} + bc - ab - ac\]\( = \left( {{a^2} - ab} \right) - \left( {ac - bc} \right)\)

\( = a\left( {a - b} \right) - c\left( {a - b} \right) = \left( {a - c} \right)\left( {a - b} \right)\)

Khi đó \(\frac{{4bc - {a^2}}}{{bc + 2{a^2}}} = \frac{{ - {{\left( {b - c} \right)}^2}}}{{\left( {a - c} \right)\left( {a - b} \right)}}\).

Tương tự, ta có: \(\frac{{4ca - {b^2}}}{{ca + 2{b^2}}} = \frac{{ - {{\left( {c - a} \right)}^2}}}{{\left( {b - a} \right)\left( {b - c} \right)}};\)

\(\frac{{4ab - {c^2}}}{{ab + 2{c^2}}} = \frac{{ - {{\left( {a - b} \right)}^2}}}{{\left( {c - a} \right)\left( {c - b} \right)}}\)

\(A = \frac{{4bc - {a^2}}}{{bc + 2{a^2}}} \cdot \frac{{4ca - {b^2}}}{{ca + 2{b^2}}} \cdot \frac{{4ab - {c^2}}}{{ab + 2{c^2}}}\)

\( = \frac{{ - {{\left( {b - c} \right)}^2}}}{{\left( {a - c} \right)\left( {a - b} \right)}} \cdot \frac{{ - {{\left( {c - a} \right)}^2}}}{{\left( {b - a} \right)\left( {b - c} \right)}} \cdot \frac{{ - {{\left( {a - b} \right)}^2}}}{{\left( {c - a} \right)\left( {c - b} \right)}} = 1\).

Lời giải

Đáp án đúng là: A

\[{\rm{A}} = \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)...\left( {1 - \frac{1}{{{n^2}}}} \right)\]

\[ = \frac{{{2^2} - 1}}{{{2^2}}} \cdot \frac{{{3^2} - 1}}{{{3^2}}} \cdot \frac{{{4^2} - 1}}{{{4^2}}} \cdot \frac{{{5^2} - 1}}{{{5^2}}} \cdots \frac{{{n^2} - 1}}{{{n^2}}}\]

\( = \frac{{1.3}}{{{2^2}}}.\frac{{2.4}}{{{3^2}}}.\frac{{3.5}}{{{4^2}}}.\frac{{4.6}}{{{5^2}}}...\frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\)

\( = \frac{{1.2.3.4...\left( {n - 1} \right)}}{{2.3.4.5...n}}.\frac{{3.4.5.6...\left( {n + 1} \right)}}{{2.3.4.5...n}}\)

\( = \frac{1}{n}.\frac{{n + 1}}{2} = \frac{{n + 1}}{{2n}}\)

Lời giải

Đáp án đúng là: C

\(A = \left( {4{x^2} - 16} \right).\frac{{7x - 2}}{{3x + 6}} = \frac{{\left( {4{x^2} - 16} \right)\left( {7x - 2} \right)}}{{3x + 6}}\)

\( = \frac{{4\left( {x - 2} \right)\left( {x + 2} \right)7x - 2}}{{3\left( {x + 2} \right)}}\)\( = \frac{{4\left( {x - 2} \right)7x - 2}}{3}\)

\( = \frac{4}{3}\left( {7{x^2} - 2x - 14x + 4} \right) = \frac{4}{3}\left( {7{x^2} - 16x + 4} \right)\)\( = \frac{4}{3}\left[ {{{\left( {\sqrt 7 x} \right)}^2} - 2.\sqrt 7 x.\frac{8}{{\sqrt 7 }} + {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2} + 4 - {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2}} \right]\)

\( = \frac{4}{3}\left[ {{{\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)}^2} - \frac{{36}}{7}} \right]\).

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

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

Dấu “=” xảy ra khi và chỉ khi \({\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)^2} = 0\) hay \(x = \frac{8}{7}\).

Lời giải

Đáp án đúng là: C

\[\left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right)\left( {1 - \frac{1}{{{5^2}}}} \right) \cdot \cdot \cdot \left( {1 - \frac{1}{{{n^2}}}} \right)\]

\[ = \frac{{{2^2} - 1}}{{{2^2}}} \cdot \frac{{{3^2} - 1}}{{{3^2}}} \cdot \frac{{{4^2} - 1}}{{{4^2}}} \cdot \frac{{{5^2} - 1}}{{{5^2}}} \cdots \frac{{{n^2} - 1}}{{{n^2}}}\]

\[ = \frac{{1.\,3}}{{{2^2}}} \cdot \frac{{2\,.\,4}}{{{3^2}}} \cdot \frac{{3\,.\,5}}{{{4^2}}} \cdot \frac{{4\,.\,6}}{{{5^2}}} \cdots \frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\]

\( = \frac{{1.2.3.4...\left( {n - 1} \right)}}{{2.3.4.5...n}}.\frac{{3.4.5.6...\left( {n + 1} \right)}}{{2.3.4.5...n}}\)

\( = \frac{1}{n}.\frac{{n + 1}}{2} = \frac{{n + 1}}{{2n}}\)

Áp dụng với n = 2010, ta có:

\[{\rm{A}} = \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right) \cdot \cdot \cdot \left( {1 - \frac{1}{{{{2010}^2}}}} \right) = \frac{{2010 + 1}}{{2\,.\,2010}} = \frac{{2011}}{{4020}}\].