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bài 6 \(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\ldots+\frac{2}{80\times82}+\frac{2}{82\times84}\)
= \(\frac12-\frac14+\frac14-\frac16+\frac16-\cdots+\frac{1}{80}-\frac{1}{82}+\frac{1}{82}-\frac{1}{84}\)
= \(\frac12-\left(\frac14-\frac14\right)-\left(\frac16-\frac16\right)-\cdots-\left(\frac{1}{82}-\frac{1}{82}\right)-\frac{1}{84}\)
=\(\frac12-0-0-\frac{1}{84}\)
=\(\frac12\) - \(\frac{1}{84}\)
=\(\frac{42}{84}-\frac{1}{84}\)
=\(\frac{41}{84}\)
Bài 7
a)\(\frac{19}{4}=4\frac34\)
b)\(\frac{27}{5}=5\frac25\)
c)\(\frac{56}{8}=7\frac08\)
a) \(\frac{-1}{4}.\frac{1}{3}=\frac{\left(-1\right).1}{4.3}\frac{-1}{12}\)
b) \(\frac{-2}{5}.\frac{5}{-9}=\frac{\left(-2\right).5}{5.\left(-9\right)}=\frac{-10}{-45}=\frac{-2}{-9}=\frac{2}{9}\)
c) \(\frac{-9}{11}.\frac{5}{18}=\frac{\left(-9\right).5}{11.18}=\frac{-45}{198}=\frac{-5}{22}\)
Kp nha!!!!
Đặt A = 9/4 + 9/28 +.. + 9/550
A = 9/1.4 + 9/4.7 +... + 9/22.25
A = 3( 3/1.4 + 3/4.7 + .. + 3/22.25)
A = 3 . (1/1 - 1/4 + 1/4 - 1/7 + ... +1/22 - 1/25)
A = 3 (1 - 1/25)
A = 3. 24 / 25
A = 72/25
Dấu "." là dấu nhân nhé
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2016.2018}\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2017}{2018}\)
\(=\frac{2017}{4036}\)
\(x.\frac{2}{7}.\frac{3}{4}=\frac{5}{21}\)
\(\frac{3}{14}x=\frac{5}{21}\)
\(x=\frac{5}{21}:\frac{3}{14}\)
\(x=\frac{5}{21}.\frac{14}{3}\)
\(x=\frac{10}{9}\)
Vậy \(x=\frac{10}{9}\)
những câu bài 2 này là tìm x bình thường
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)