Bài 10: Phép nhân phân số

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Đặt A= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)

= \(\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\right)\)

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}......1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5.....9.11}\) ( Bước này bạn bỏ đi cũng được )

\(=\dfrac{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}\)

\(=\dfrac{\left(2.3.4.....9\right).10.2.\left(3.4.5.....10\right)}{1.\left(2.3.4.....9\right).\left(3.4.5.....10\right).11}\)

\(=\dfrac{10.2}{1.11}=\dfrac{20}{11}=1\dfrac{9}{11}\)

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5....9.11}\)

\(=\dfrac{2.3.4....10}{1.2.3....9}.\dfrac{2.3.4...10}{3.4.5....11}\)

\(=10.\dfrac{2}{11}=\dfrac{20}{11}\)

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.3.4.....10}{1.2.3.....9}.\dfrac{2.3.4.....10}{3.4.5.....11}\)

\(=10.\dfrac{2}{11}\)

\(=\dfrac{20}{11}\)

b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)

Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé

Sai đề ak

Ta có:

\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right).x=1\)

\(\Leftrightarrow\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}.x=1\)

\(\Leftrightarrow\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.....\dfrac{1558}{1560}.x=1\)

\(\Leftrightarrow\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}.......\dfrac{38.41}{39.40}.x=1\)

\(\Leftrightarrow\dfrac{1.4.2.5.3.6.....38.41}{2.3.3.4.4.5.....39.40}.x=1\)

\(\Leftrightarrow\dfrac{\left(1.2.3.4....38\right)\left(4.5.6.7....41\right)}{\left(2.3.4.....39\right)\left(3.4.5....40\right)}.x=1\)

\(\Leftrightarrow\dfrac{1}{39}.\dfrac{41}{3}.x=1\)

\(\Leftrightarrow\dfrac{41}{117}.x=1\)

\(\Leftrightarrow x=\dfrac{117}{41}\)

Vậy ...

Thanhs bạn nha

Ta có : \(\dfrac{-1}{2}\).\(\dfrac{-2}{3}\)\(\dfrac{-3}{4}\). ... . \(\dfrac{-98}{99}\)

=>\(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right).....\left(-98\right)}{2.3.4.....99}\)

=> \(\dfrac{1}{99}\)

Tick mình nha bạn hiền.

(1/2-1) (1/3-1) (1/4-1) ... (1/99-1)

=(-1/2) (-2/3) (-3/4) ... (-98/99)

=(-1/99)

Vậy (1/2-1) (1/3-1) (1/4-1) ... (1/99-1)=(-1/99)

Đặt \(A\) \(=\) \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{99}-1\right)\)

\(A=\left(-\dfrac{1}{2}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{3}{4}\right)...\left(-\dfrac{98}{99}\right)\)

\(A=\dfrac{1.2.3.....98}{2.3.4.....99}\)

\(A=\dfrac{1}{99}\)

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