[SỐ HỌC 6] BÀI TẬP TOÁN NÂNG CAO - PHẦN 1 (CÓ ĐÁP ÁN)

Bài toán 1: Tìm $n$ $ \in $ $\mathbb{N}$ để ba phân số $\dfrac{{21}}{n}$, $\dfrac{{22}}{{n - 1}}$, $\dfrac{{24}}{{n + 1}}$ đều là số tự nhiên.

Giải.

Để $\dfrac{{21}}{a}$ là số tự nhiên thì $a = 3; 7; 21$.

Vậy để cả ba phân số $\dfrac{{21}}{a}$, $\dfrac{{22}}{{a - 1}}$, $\dfrac{{24}}{{a + 1}}$ đều là số tự nhiên thì $a$ chỉ có thể là $3$; $7$ hoặc $21$.

Khi $a = 3$ thì $\dfrac{{21}}{a} = \dfrac{{21}}{3} = 7$; $\dfrac{{22}}{{a - 1}} = \dfrac{{22}}{{3 - 1}} = \dfrac{{22}}{2} = 11$; $\dfrac{{24}}{{a + 1}} = \dfrac{{24}}{{3 + 1}} = \dfrac{{24}}{4} = 6$ thỏa mãn yêu cầu bài toán.

Khi $a = 7$ thì $\dfrac{{21}}{a} = \dfrac{{21}}{7} = 3$; $\dfrac{{22}}{{a - 1}} = \dfrac{{22}}{{7 - 1}} = \dfrac{{22}}{6}$ không phải là số tự nhiên.

Khi $a = 21$ thì $\dfrac{{21}}{a} = \dfrac{{21}}{{21}} = 1$; $\dfrac{{22}}{{a - 1}} = \dfrac{{22}}{{21 - 1}} = \dfrac{{22}}{{20}}$ không phải là số tự nhiên.

Vậy $a = 3$ thì cả ba phân số đã cho là số tự nhiên.
Bài toán 2: Tìm $n$ $ \in $ $\mathbb{Z}$ để $A = \dfrac{{11}}{{n + 1}}$, $B = \dfrac{{17}}{{n - 3}}$ là số nguyên.

Giải.

Để $A$ là số nguyên thì $11\; \vdots \;\left( {n + 1} \right)$
$ \Rightarrow $ $n + 1 \in $ Ư(11)
$ \Rightarrow $ $n + 1 \in \left\{ {1;11; - 1; - 11} \right\}$
$ \Rightarrow $ $n \in \left\{ {0;10; - 2; - 12} \right\}$
Để $B$ là số nguyên thì $17\; \vdots \;\left( {n - 3} \right)$
$ \Rightarrow $ $n - 3 \in $ Ư(17)
$ \Rightarrow $ $n - 3 \in \left\{ {1;17; - 1; - 17} \right\}$
$ \Rightarrow $ $n \in \left\{ {4; 21; 2; - 14} \right\}$
Bài toán 3: Chứng tỏ rằng $\dfrac{{21n + 4}}{{14n + 3}}$ là phân số tối giản ($n$ $ \in $ $\mathbb{N}$).

Giải.

Đặt ƯCLN($12n + 4$, $14n + 3$) = $d$ ($d > 0$).
$ \Rightarrow $ $14n + 3\; \vdots \;d$, $21n + 4\; \vdots \;d$
$ \Rightarrow $ $3(14n + 3)\; \vdots \;d$, $2(21n + 4)\; \vdots \;d$
$ \Rightarrow $ $\left[ {3\left( {14n + 3} \right) - 2\left( {21n + 4} \right)} \right]\; \vdots \;d$
$ \Rightarrow $ $\left[ {\left( {42n + 9} \right) - \left( {42n + 8} \right)} \right]\; \vdots \;d$
$ \Rightarrow $ $1\; \vdots \;d$
$ \Rightarrow $ $d = 1$
$ \Rightarrow $ $21n + 4$ và $14n + 3$ là hai số nguyên tố cùng nhau.
$ \Rightarrow $ $\dfrac{{21n + 4}}{{14n + 3}}$ là phân số tối giản.
Bài toán 4: Tìm tất cả các số tự nhiên $n$ để $\dfrac{{5n + 6}}{{6n + 5}}$ có thể rút gọn được.

Giải.

Đặt ƯCLN($5n + 6$, $6n + 5$) = $d$ ($d > 0$).
$ \Rightarrow $ $5n + 6\; \vdots \;d$, $6n + 5\; \vdots \;d$
$ \Rightarrow $ $6(5n + 6)\; \vdots \;d$, $5(6n + 5)\; \vdots \;d$
$ \Rightarrow $ $\left[ {6\left( {5n + 6} \right) - 5\left( {6n + 5} \right)} \right]\; \vdots \;d$
$ \Rightarrow $ $\left[ {\left( {30n + 36} \right) - \left( {30n + 25} \right)} \right]\; \vdots \;d$
$ \Rightarrow $ $11\; \vdots \;d$
$ \Rightarrow $ $d \in \left\{ {1;11} \right\}$
Vì $\dfrac{{5n + 6}}{{6n + 5}}$ rút gọn được nên $d = 11$.
Ta có:
$5n + 6\; \vdots \;11$, $6n + 5\; \vdots \;11$
$ \Rightarrow $ $\left[ {\left( {6n + 5} \right) - \left( {5n + 6} \right)} \right]\; \vdots \;11$
$ \Rightarrow $ $n - 1\; \vdots \;11$
$ \Rightarrow $ $n - 1 = 11k$ $\left( {k \ge 0} \right)$
$ \Rightarrow $ $n = 11k + 1$ $\left( {k \ge 0} \right)$

Bài toán 5: Với $n$ $ \in $ $\mathbb{N}$$^*$, chứng tỏ rằng:

a) $\dfrac{1}{{n(n + 1)}} = \dfrac{1}{n} - \dfrac{1}{{n + 1}}$

b) $\dfrac{2}{{n(n + 1)(n + 2)}} = \dfrac{1}{{n(n + 1)}} - \dfrac{1}{{(n + 1)(n + 2)}}$

Giải.

a) $\dfrac{1}{{n(n + 1)}} = \dfrac{{(n + 1) - n}}{{n(n + 1)}} = \dfrac{{n + 1}}{{n(n + 1)}} - \dfrac{n}{{n(n + 1)}} = \dfrac{1}{n} - \dfrac{1}{{n + 1}}$

b) $\dfrac{2}{{n(n + 1)(n + 2)}} = \dfrac{{(n + 2) - n}}{{n(n + 1)(n + 2)}} = \dfrac{1}{{n(n + 1)}} - \dfrac{1}{{(n + 1)(n + 2)}}$

Bài toán 6: Chứng tỏ rằng:

a) $\dfrac{1}{{{n^2}}} < \dfrac{1}{{n - 1}} - \dfrac{1}{n}$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 2$)

b) $\dfrac{1}{{{n^2}}} > \dfrac{1}{{n}} - \dfrac{1}{n + 1}$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 1$)

Giải.

a) $\dfrac{1}{{{n^2}}} = \dfrac{1}{{n \cdot n}} < \dfrac{1}{{(n - 1)n}} = \dfrac{1}{{n - 1}} - \dfrac{1}{n}$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 2$)

b) $\dfrac{1}{{{n^2}}} = \dfrac{1}{{n \cdot n}} > \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{1}{n} - \dfrac{1}{{n + 1}}$ ($n$ $ \in $ $\mathbb{N}$, $n \ge 1$)

Bài toán 7: Cho $a$, $b$, $m$ $ \in $ $\mathbb{N}$$^*$. Chứng tỏ rằng:

a) $\dfrac{a}{b} > \dfrac{{a + m}}{{b + m}}$ khi $a > b$
b) $\dfrac{a}{b} < \dfrac{{a + m}}{{b + m}}$ khi $a < b$

Giải.

a) $a > b$ $ \Rightarrow $ $a - b > 0$

$\dfrac{a}{b} = \dfrac{{b + a - b}}{b} = 1 + \dfrac{{a - b}}{b}$

$\dfrac{{a + m}}{{b + m}} = \dfrac{{(b + m) + (a - b)}}{{b + m}} = 1 + \dfrac{{a - b}}{{b + m}}$

$\dfrac{{a - b}}{b} > \dfrac{{a - b}}{{b + m}}$ (do $a, b, m > 0$, $a - b > 0$)

$ \Rightarrow $ $1 + \dfrac{{a - b}}{b} > 1 + \dfrac{{a - b}}{{b + m}}$

$ \Rightarrow $ $\dfrac{a}{b} > \dfrac{{a + m}}{{b + m}}$

b) Tương tự

Bài toán 8: Tính các tổng sau đây:

1) $A = \dfrac{1}{{1\cdot2}} + \dfrac{1}{{2\cdot3}} + \dfrac{1}{{3\cdot4}} +  \cdots  + \dfrac{1}{{99\cdot100}}$

2) $B = \dfrac{5}{{1\cdot6}} + \dfrac{5}{{6\cdot11}} + \dfrac{5}{{11\cdot16}} + ... + \dfrac{5}{{96\cdot101}}$

3) $C = \dfrac{1}{{1\cdot5}} + \dfrac{1}{{5\cdot9}} + \dfrac{1}{{9\cdot13}} +  \cdots  + \dfrac{1}{{37\cdot41}}$

4) $D = \dfrac{{11}}{{1\cdot3}} + \dfrac{{11}}{{3\cdot5}} + \dfrac{{11}}{{5\cdot7}} +  \cdots  + \dfrac{{11}}{{97\cdot99}}$

5) $E = \dfrac{1}{6} + \dfrac{1}{{10}} + \dfrac{1}{{15}} + \dfrac{1}{{21}} + \dfrac{1}{{28}} + \dfrac{1}{{36}} + \dfrac{1}{{45}} + \dfrac{1}{{55}} + \dfrac{1}{{66}} + \dfrac{1}{{78}} + \dfrac{1}{{91}} + \dfrac{1}{{105}}$

6) $F = \dfrac{1}{{1\cdot2\cdot3}} + \dfrac{1}{{2\cdot3\cdot4}} + \dfrac{1}{{3\cdot4\cdot5}} +  \cdots  + \dfrac{1}{{98\cdot99\cdot100}}$

7) $G = \dfrac{1}{{2\cdot7\cdot12}} + \dfrac{1}{{7\cdot12\cdot17}} + \dfrac{1}{{12\cdot17\cdot22}} +  \cdots  + \dfrac{1}{{1997\cdot2002\cdot2007}}$
8) $H = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2004}}}}$
9) $I = \dfrac{7}{{10}} + \dfrac{7}{{{{10}^2}}} + \dfrac{7}{{{{10}^3}}} +  \ldots $

10) $J = 1 + \dfrac{1}{2}\left( {1 + 2} \right) + \dfrac{1}{3}\left( {1 + 2 + 3} \right) + \dfrac{1}{4}\left( {1 + 2 + 3 + 4} \right) +  \cdots  + $

$ + \dfrac{1}{{16}}\left( {1 + 2 + 3 +  \cdots  + 16} \right)$

Giải.

1) $A = \dfrac{1}{{1\cdot2}} + \dfrac{1}{{2\cdot3}} + \dfrac{1}{{3\cdot4}} +  \cdots  + \dfrac{1}{{99\cdot100}}$

= $\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} +  \cdots  + \dfrac{1}{{99}} - \dfrac{1}{{100}}$

= $\dfrac{1}{1} - \dfrac{1}{{100}}$

= $\dfrac{{99}}{{100}}$

2) $B = \dfrac{5}{{1\cdot6}} + \dfrac{5}{{6\cdot11}} + \dfrac{5}{{11\cdot16}} + ... + \dfrac{5}{{96\cdot101}}$

= $\dfrac{1}{1} - \dfrac{1}{6} + \dfrac{1}{6} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{16}} + ... + \dfrac{1}{{96}} - \dfrac{1}{{101}}$

= $\dfrac{1}{1} - \dfrac{1}{{101}}$

= $\dfrac{{100}}{{101}}$

3) $C = \dfrac{1}{{1\cdot5}} + \dfrac{1}{{5\cdot9}} + \dfrac{1}{{9\cdot13}} +  \cdots  + \dfrac{1}{{37\cdot41}}$

= $\dfrac{1}{4}\left( {\dfrac{4}{{1 \cdot 5}} + \dfrac{4}{{5 \cdot 9}} + \dfrac{4}{{9 \cdot 13}} +  \cdots  + \dfrac{4}{{37 \cdot 41}}} \right)$

= $\dfrac{1}{4}\left( {\dfrac{1}{1} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{{13}} +  \cdots  + \dfrac{1}{{37}} - \dfrac{1}{{41}}} \right)$

= $\dfrac{1}{4}\left( {\dfrac{1}{1} - \dfrac{1}{{41}}} \right)$

= $\dfrac{{10}}{{41}}$

4) $D = \dfrac{{11}}{{1 \cdot 3}} + \dfrac{{11}}{{3 \cdot 5}} + \dfrac{{11}}{{5 \cdot 7}} +  \cdots  + \dfrac{{11}}{{97 \cdot 99}}$

= $\dfrac{{11}}{2}\left( {\dfrac{2}{{1 \cdot 3}} + \dfrac{2}{{3 \cdot 5}} + \dfrac{2}{{5 \cdot 7}} +  \cdots  + \dfrac{2}{{97 \cdot 99}}} \right)$

= $\dfrac{{11}}{2}\left( {\dfrac{1}{1} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} +  \cdots  + \dfrac{1}{{97}} - \dfrac{1}{{99}}} \right)$

= $\dfrac{{11}}{2}\left( {\dfrac{1}{1} - \dfrac{1}{{99}}} \right)$

= $\dfrac{{49}}{9}$

5) $E = \dfrac{1}{6} + \dfrac{1}{{10}} + \dfrac{1}{{15}} + \dfrac{1}{{21}} + \dfrac{1}{{28}} + \dfrac{1}{{36}} + \dfrac{1}{{45}} + \dfrac{1}{{55}} + \dfrac{1}{{66}} + \dfrac{1}{{78}} + \dfrac{1}{{91}} + \dfrac{1}{{105}}$

= $\dfrac{2}{{12}} + \dfrac{2}{{20}} + \dfrac{2}{{30}} + \dfrac{2}{{42}} + \dfrac{2}{{56}} + \dfrac{2}{{72}} + \dfrac{2}{{90}} + \dfrac{2}{{110}} + \dfrac{2}{{132}} + \dfrac{2}{{156}} + \dfrac{2}{{182}} + \dfrac{2}{{210}}$

= $2\left( {\dfrac{1}{{3 \cdot 4}} + \dfrac{1}{{4 \cdot 5}} + \dfrac{1}{{5 \cdot 6}} +  \cdots  + \dfrac{1}{{13 \cdot 14}} + \dfrac{1}{{14 \cdot 15}}} \right)$

= (làm tương tự như các câu trên)

= $\dfrac{{8}}{{15}}$

6) $F = \dfrac{1}{{1 \cdot 2 \cdot 3}} + \dfrac{1}{{2 \cdot 3 \cdot 4}} + \dfrac{1}{{3 \cdot 4 \cdot 5}} +  \cdots  + \dfrac{1}{{98 \cdot 99 \cdot 100}}$

= $\dfrac{1}{2}\cdot\dfrac{2}{{1 \cdot 2 \cdot 3}} + \dfrac{1}{2}\cdot\dfrac{2}{{2 \cdot 3 \cdot 4}} + \dfrac{1}{2}\cdot\dfrac{2}{{3 \cdot 4 \cdot 5}} +  \cdots  + \dfrac{1}{2}\cdot\dfrac{2}{{98 \cdot 99 \cdot 100}}$
= $\dfrac{1}{2}\cdot\left( {\dfrac{2}{{1 \cdot 2 \cdot 3}} + \dfrac{2}{{2 \cdot 3 \cdot 4}} + \dfrac{2}{{3 \cdot 4 \cdot 5}} +  \cdots  + \dfrac{2}{{98 \cdot 99 \cdot 100}}} \right)$
= $\dfrac{1}{2}\left( {\dfrac{1}{{1 \cdot 2}} - \dfrac{1}{{2 \cdot 3}} + \dfrac{1}{{2 \cdot 3}} - \dfrac{1}{{3 \cdot 4}} + \dfrac{1}{{3 \cdot 4}} - \dfrac{1}{{4 \cdot 5}} +  \cdots  + \dfrac{1}{{98 \cdot 99}} - \dfrac{1}{{99 \cdot 100}}} \right)$
= $\dfrac{1}{2}\left( {\dfrac{1}{{1 \cdot 2}} - \dfrac{1}{{99 \cdot 100}}} \right)$
= $\dfrac{{4949}}{{19800}}$

7) $G = \dfrac{1}{{2 \cdot 7 \cdot 12}} + \dfrac{1}{{7 \cdot 12 \cdot 17}} + \dfrac{1}{{12 \cdot 17 \cdot 22}} +  \cdots  + \dfrac{1}{{1997 \cdot 2002 \cdot 2007}}$

= $\dfrac{1}{{10}}\left( {\dfrac{{12 - 2}}{{2 \cdot 7 \cdot 12}} + \dfrac{{17 - 7}}{{7 \cdot 12 \cdot 17}} + \dfrac{{22 - 12}}{{12 \cdot 17 \cdot 22}} +  \cdots  + \dfrac{{2007 - 1997}}{{1997 \cdot 2002 \cdot 2007}}} \right)$

= $\dfrac{1}{{10}}\left( {\dfrac{1}{{2 \cdot 7}} - \dfrac{1}{{7 \cdot 12}} + \dfrac{1}{{7 \cdot 12}} - \dfrac{1}{{12 \cdot 17}} + \dfrac{1}{{12 \cdot 17}} - \dfrac{1}{{17 \cdot 22}} +  \cdots } \right. + \dfrac{1}{{1997 \cdot 2002}}$

$\left. { - \dfrac{1}{{2002 \cdot 2007}}} \right)$

= $\dfrac{1}{{10}}\left( {\dfrac{1}{{2 \cdot 7}} - \dfrac{1}{{2002 \cdot 2007}}} \right)$

= $\dfrac{{71}}{{20090070}}$

8) $H = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2004}}}}$

$ \Rightarrow $ $H + \dfrac{1}{{{2^{2004}}}} = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} +  \ldots  + \dfrac{1}{{{2^{2004}}}} + \dfrac{1}{{{2^{2004}}}}$

$ \Rightarrow $ $H + \dfrac{1}{{{2^{2004}}}} = 1$

$ \Rightarrow $ $H = 1 - \dfrac{1}{{{2^{2004}}}} = \dfrac{{{2^{2004}} - 1}}{{{2^{2004}}}}$

9) $I = \dfrac{7}{{10}} + \dfrac{7}{{{{10}^2}}} + \dfrac{7}{{{{10}^3}}} +  \ldots $

$I - \dfrac{7}{{10}} = \dfrac{7}{{{{10}^2}}} + \dfrac{7}{{{{10}^3}}} +  \ldots $
$10\left( {I - \dfrac{7}{{10}}} \right) = 10\left( {\dfrac{7}{{{{10}^2}}} + \dfrac{7}{{{{10}^3}}} +  \ldots } \right) = \dfrac{7}{{10}} + \dfrac{7}{{{{10}^2}}} +  \ldots  = I$
$10I - 7 = I$
$9I = 7$
$I = \dfrac{7}{9}$

10) $J = 1 + \dfrac{1}{2}\left( {1 + 2} \right) + \dfrac{1}{3}\left( {1 + 2 + 3} \right) + \dfrac{1}{4}\left( {1 + 2 + 3 + 4} \right) +  \cdots  + $

$ + \dfrac{1}{{16}}\left( {1 + 2 + 3 +  \cdots  + 16} \right)$

= $1 + \dfrac{1}{2}\cdot\dfrac{{2 \cdot 3}}{2} + \dfrac{1}{3}\cdot\dfrac{{3 \cdot 4}}{2} + \dfrac{1}{4}\cdot\dfrac{{4 \cdot 5}}{2} +  \cdots  + \dfrac{1}{{16}}\cdot\dfrac{{16 \cdot 17}}{2}$
= $1 + \dfrac{3}{2} + \dfrac{4}{2} + \dfrac{5}{2} +  \cdots  + \dfrac{{17}}{2}$
= $\dfrac{1}{2}\left( {2 + 3 + 4 + 5 +  \cdots  + 17} \right)$
= $\dfrac{1}{2}\left( {\dfrac{{17 \cdot 18}}{2} - 1} \right)$
= $\dfrac{1}{2}\cdot152$
= $76$