Lời giải

Đáp án đúng là: B

Phân thức nghịch đảo của phân thức $\frac{{2{\rm{x}} + 1}}{{{\rm{x}} + 2}}$ là $\frac{{{\rm{x}} + 2}}{{2{\rm{x}} + 1}}$.

Lời giải

Đáp án đúng là: D

$\frac{{3{\rm{x}} + 12}}{{4{\rm{x}} - 16}} \cdot \frac{{8 - 2{\rm{x}}}}{{{\rm{x}} + 4}} = \frac{{3\left( {{\rm{x}} + 4} \right)}}{{4\left( {{\rm{x}} - 4} \right)}} \cdot \frac{{2\left( {4 - {\rm{x}}} \right)}}{{{\rm{x}} + 4}}$

$= \frac{{3\left( {{\rm{x}} + 4} \right)}}{{4\left( {{\rm{x}} - 4} \right)}} \cdot \frac{{ - 2\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} + 4}} = \frac{{ - 3}}{2}$.

Lời giải

Đáp án đúng là: C

$\frac{{4{\rm{x}} + 12}}{{{{\left( {{\rm{x}} + 4} \right)}^2}}}:\frac{{3\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 4}} = \frac{{4\left( {{\rm{x}} + 3} \right)}}{{{{\left( {{\rm{x}} + 4} \right)}^2}}}:\frac{{3\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 4}}$

$= \frac{{4\left( {{\rm{x}} + 3} \right)}}{{{{\left( {{\rm{x}} + 4} \right)}^2}}} \cdot \frac{{{\rm{x}} + 4}}{{3\left( {{\rm{x}} + 3} \right)}} = \frac{4}{{3\left( {{\rm{x}} + 4} \right)}}$.

Lời giải

Đáp án đúng là: A

$\frac{{{{\rm{x}}^3} + 1}}{{{{\rm{x}}^2} + 2{\rm{x}} + 1}}:\frac{{3{{\rm{x}}^2} - 3{\rm{x}} + 3}}{{{{\rm{x}}^2} - 1}}$

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

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

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

Vậy kết quả của phép chia $\frac{{{{\rm{x}}^3} + 1}}{{{{\rm{x}}^2} + 2{\rm{x}} + 1}}:\frac{{3{{\rm{x}}^2} - 3{\rm{x}} + 3}}{{{{\rm{x}}^2} - 1}}$ có tử thức là x − 1.

Lời giải

Đáp án đúng là: A

$\frac{{{\rm{x + 3}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{4}}}}{\rm{.}}\frac{{{\rm{8}} - {\rm{12x + 6}}{{\rm{x}}^{\rm{2}}} - {{\rm{x}}^{\rm{3}}}}}{{{\rm{9x + 27}}}}$

${\rm{ = }}\frac{{{\rm{x + 3}}}}{{\left( {{\rm{x}} - {\rm{2}}} \right)\left( {{\rm{x + 2}}} \right)}}{\rm{.}}\frac{{{{\left( {{\rm{2}} - {\rm{x}}} \right)}^{\rm{3}}}}}{{{\rm{9}}\left( {{\rm{x + 3}}} \right)}}$

${\rm{ = }} - \frac{{{{\left( {{\rm{x}} - {\rm{2}}} \right)}^{\rm{2}}}}}{{{\rm{9}}\left( {{\rm{x + 2}}} \right)}}$.

Vậy${\rm{A = }}{\left( {{\rm{x}} - {\rm{2}}} \right)^{\rm{2}}}{\rm{;}}\,\,{\rm{B = 9}}\left( {{\rm{x + 2}}} \right)$.

Lời giải

Đáp án đúng là: B

$A:\frac{{x + 1}}{{{x^2} + x + 1}} = \frac{{{x^3} - 1}}{{{x^2} - 1}}$

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

Lời giải

Đáp án đúng là: D

$\frac{{x + 3y}}{{4x + 8y}}\,\,.\,\,A = \frac{{{x^2} - 9{y^2}}}{{x + 2y}}$

$A = \frac{{{x^2} - 9{y^2}}}{{x + 2y}}:\frac{{x + 3y}}{{4x + 8y}}$

$= \frac{{\left( {x - 3y} \right)\left( {x + 3y} \right)}}{{x + 2y}}:\frac{{x + 3y}}{{4\left( {x + 2y} \right)}}$

$= \frac{{\left( {x - 3y} \right)\left( {x + 3y} \right)}}{{x + 2y}} \cdot \frac{{4\left( {x + 2y} \right)}}{{x + 3y}} = 4\left( {x - 3y} \right)$.

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:

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