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\(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{10}\right).\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}\)
\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(=\dfrac{4.10.18...1558}{6.12.20...1560}\)
\(=\dfrac{41}{39}.3\)
\(=\dfrac{41}{11}\)
\(B=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(B=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(B=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}.....\dfrac{38.41}{39.40}\)
\(B=\dfrac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\dfrac{1.41}{39.3}=\dfrac{41}{117}\)
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 ...
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
2.
\(\dfrac{a}{3}-\dfrac{2}{b}=\dfrac{1}{3}\)
\(\dfrac{a\times b-3\times2}{3\times b}\)\(=\dfrac{1}{3}\)
\(\dfrac{a\times b-6}{3\times b}=\dfrac{1}{3}\)
\(\Rightarrow3\times\left(a\times b-6\right)=1\times\left(3\times b\right)\)
\(3ab-18=3b\)
\(3ab-18-3b=0\)
\(3ab-3b=18\)
\(3b\left(a-1\right)=18\)
Mà \(18=1.18=2.9=3.6\)
\(\Rightarrow3b\left(a-1\right)=1.18=2.9=3.6\)
còn lại bạn tự làm các trường hợp ra nhé,mk lười lắm
\(T=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{10}\right)\)\(\Rightarrow T=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
\(\Rightarrow=\dfrac{1}{11}\)
\(\Rightarrow\) Số nghịch đảo của T là \(11\)
\(A=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)......\left(1-\dfrac{1}{780}\right)\)
= \(\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.......\dfrac{779}{780}\)
= \(\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.....\dfrac{1558}{1560}\)
= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}......\dfrac{38.41}{39.40}\)
= \(\dfrac{1.4.2.5.3.6.....38.41}{2.3.3.4.4.5...39.40}\)
= \(\dfrac{\left(1.2.3....38\right)\left(4.5.6....41\right)}{\left(2.3.4....39\right)\left(3.4.5...40\right)}\)
= \(\dfrac{1}{39}.\dfrac{41}{3}\) = \(\dfrac{41}{117}\)
làm đúng rồi đó, Thái cũng làm như mi