\(\frac{13}{14}\)\(-\)\(\frac{3}{12}\) \(\frac{21}{7}\)\(-\)2
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a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
\(A=\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{9901}{9900}=\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+\left(1+\frac{1}{4.5}\right)+...+\left(1+\frac{1}{99.100}\right)\)\(=\left(1+1+1+...+1\right)+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=98+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98+\frac{49}{100}=98\frac{49}{100}\)
a) $\frac{4}{{25}}:\frac{4}{3} = \frac{4}{{25}} \times \frac{3}{4} = \frac{3}{{25}}$
b) $\frac{3}{{14}}:\frac{6}{7} = \frac{3}{{14}} \times \frac{7}{6} = \frac{{3 \times 7}}{{14 \times 6}} = \frac{{3 \times 7}}{{7 \times 2 \times 3 \times 2}} = \frac{1}{4}$
c) $\frac{{12}}{{15}}:2 = \frac{{12}}{{15}} \times \frac{1}{2} = \frac{{12 \times 1}}{{15 \times 2}} = \frac{{6 \times 2 \times 1}}{{15 \times 2}} = \frac{6}{{15}}$
d) $\frac{{21}}{8}:6 = \frac{{21}}{8} \times \frac{1}{6} = \frac{{21 \times 1}}{{8 \times 6}} = \frac{{7 \times 3 \times 1}}{{8 \times 3 \times 2}} = \frac{7}{{16}}$
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
Gọi tổng dãy số hạng trên là A
A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)
Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số
A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\)
A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)
\(\left(1+2\frac{1}{4}-3\frac{1}{3}\right):\left(1+3\frac{1}{2}-4\frac{7}{12}\right)\)
\(=\left(1+2-3+\frac{1}{4}-\frac{1}{3}\right):\left(1+3-4+\frac{1}{2}-\frac{7}{12}\right)\)
\(=\frac{-1}{12}:\frac{-1}{12}=\frac{-1}{12}.\frac{12}{-1}=\frac{12}{12}=1\)
a. (1 + \(2\frac{1}{4}\) - \(3\frac{1}{3}\)) : (1 + \(3\frac{1}{2}\) - \(4\frac{7}{12}\))
= \(-\frac{1}{12}\): ( \(-\frac{1}{12}\))
= 1
b. \(5\frac{1}{2}\)- \(14\frac{3}{7}\) : \(\frac{9}{13}\)- \(3\frac{4}{7}\) : \(\frac{9}{13}\)
= \(\frac{11}{2}\)+ (\(14\frac{3}{7}-3\frac{4}{7}\)) : \(\frac{9}{13}\)
= \(\frac{11}{2}\)- 26
= \(-\frac{41}{2}\)
k mk đi
ai k mk
mk lại
thanks
\(\frac{13}{14}\) - \(\frac{3}{12}\) = \(\frac{78}{84}\) - \(\frac{21}{84}\) = \(\frac{57}{84}\) = \(\frac{19}{28}\)
\(\frac{21}{7}\)- 2 = \(\frac{21}{7}\) - \(\frac{14}{7}\) = 1