5 bài tập Tính nhanh tính thuận tiện với phân số

Hướng Dẫn Giải

a) Ta có: m = 1; n = 1; a = 2; z = 20. Áp dụng công thức tính nhanh ta có kết quả:

$\frac{1}{1} \times (\frac{1}{2} - \frac{1}{{20}}) = \frac{9}{{20}}$

b) Viết lại: $\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + ..... + \frac{1}{{72}} + \frac{1}{{90}} = \frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + ..... + \frac{1}{{8 \times 9}} + \frac{1}{{9 \times 10}}$

Áp dụng công thức tính nhanh với: m = n = 1; a = 1; z = 10 được kết quả:

$\frac{1}{1} \times (\frac{1}{1} - \frac{1}{{10}}) = 1 - \frac{1}{{10}} = \frac{9}{{10}}$

Đáp Số: a) $\frac{9}{{20}}$; b) $\frac{9}{{10}}$.

Hướng Dẫn Giải

Đặt $A = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}}$

Suy ra: $3 \times A = 3 \times (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}})$

$= 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}$.

Do đó:

$3 \times A - A = (3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}) - (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}} + \frac{1}{{729}})$

$\to 2 \times A = 3 - \frac{1}{{729}} = \frac{{2187}}{{729}} - \frac{1}{{729}} = \frac{{2186}}{{729}}$

$\to A = \frac{{2186}}{{729}}:2 = \frac{{1093}}{{729}}$

Đáp Số: $A = \frac{{1093}}{{729}}$.

Hướng Dẫn Giải

$(1 - \frac{3}{4}) \times (1 - \frac{3}{7}) \times (1 - \frac{3}{{10}}) \times \ldots \times (1 - \frac{3}{{100}}) = \ldots$

$= \frac{1}{4} \times \frac{4}{7} \times \frac{7}{{10}} \times \ldots \times \frac{{97}}{{100}} = \frac{1}{{100}}$

Đáp Số: $\frac{1}{{100}}$

Hướng Dẫn Giải

Ta tính:

$\frac{1}{{1 \times 99}} + \frac{1}{{3 \times 97}} + \frac{1}{{5 \times 95}} + \ldots + \frac{1}{{97 \times 3}} + \frac{1}{{99 \times 1}}$

$= \frac{2}{{1 \times 99}} + \frac{2}{{3 \times 97}} + \frac{2}{{5 \times 95}} \ldots + \frac{2}{{49 \times 51}}$

$= \frac{1}{{50}} \times (\frac{{50 \times 2}}{{1 \times 99}} + \frac{{50 \times 2}}{{3 \times 97}} + \frac{{50 \times 2}}{{5 \times 95}} + \ldots + \frac{{50 \times 2}}{{49 \times 51}})$

$= \frac{1}{{50}} \times (\frac{{100}}{{1 \times 99}} + \frac{{100}}{{3 \times 97}} + \frac{{100}}{{5 \times 95}} + \ldots + \frac{{100}}{{49 \times 51}})$

$= \frac{1}{{50}} \times (\frac{{1 + 99}}{{1 \times 99}} + \frac{{3 + 97}}{{3 \times 97}} + \frac{{5 + 95}}{{5 \times 95}} + \ldots + \frac{{49 + 51}}{{49 \times 51}})$

$= \frac{1}{{50}} \times \left( {\frac{1}{{1 \times 99}} + \frac{{99}}{{1 \times 99}} + \frac{3}{{3 \times 97}} + \frac{{97}}{{3 \times 97}} + \frac{5}{{5 \times 95}} + \frac{{95}}{{5 \times 95}} + \ldots + \frac{{49}}{{49 \times 51}} + \frac{{51}}{{49 \times 51}}} \right)$

$= \frac{1}{{50}} \times \left( {\frac{1}{{99}} + 1 + \frac{1}{{97}} + \frac{1}{3} + \frac{1}{{95}} + \frac{1}{5} + \ldots + \frac{1}{{49}} + \frac{1}{{51}}} \right)$

$= \frac{1}{{50}} \times (1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{97}} + \frac{1}{{99}})$

Vậy:

$\frac{{1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{97}} + \frac{1}{{99}}}}{{\frac{1}{{1 \times 99}} + \frac{1}{{3 \times 97}} + \frac{1}{{5 \times 95}} + \ldots + \frac{1}{{97 \times 3}} + \frac{1}{{99 \times 1}}}} = \frac{{1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{97}} + \frac{1}{{99}}}}{{\frac{1}{{50}} \times (1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{97}} + \frac{1}{{99}})}} = \frac{1}{{\frac{1}{{50}}}} = 50$Đáp Số: 50.

Hướng Dẫn Giải

Tử số = $\frac{{2017}}{1} + \frac{{2016}}{2} + \frac{{2015}}{3} + \ldots + \frac{1}{{2017}} + 2017$

$= 2017 + \frac{{2016}}{2} + \frac{{2015}}{3} + \ldots + \frac{1}{{2017}} + \underbrace {1 + 1 + \ldots + 1}_{}$

2017 số 1

$= (1 + 2017) + (1 + \frac{{2016}}{2}) + (1 + \frac{{2015}}{3}) + \ldots + (1 + \frac{1}{{2017}})$

$= 2018 + \frac{{2018}}{2} + \frac{{2018}}{3} + \ldots + \frac{{2018}}{{2017}}$

$= 2018 \times (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{2017}})$

Do đó:

$A = \frac{{2018 \times (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{2017}})}}{{1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{2017}}}} = 2018$.