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a) $\frac{{14}}{{18}}:\frac{8}{9} = \frac{7}{9}:\frac{8}{9} = \frac{7}{9} \times \frac{9}{8} = \frac{{63}}{{72}} = \frac{7}{8}$
b) $\frac{9}{6}:\frac{3}{{10}} = \frac{3}{2}:\frac{3}{{10}} = \frac{3}{2} \times \frac{{10}}{3} = \frac{{30}}{6} = 5$
c) $\frac{4}{5}:\frac{{10}}{{15}} = \frac{4}{5}:\frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{{12}}{{10}} = \frac{6}{5}$
d) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
Ta có: \(\frac56=1-\frac16;\frac78=1-\frac18;\frac{9}{10}=1-\frac{1}{10}\)
Ta có: 6<8<10
=>\(\frac16>\frac18>\frac{1}{10}\)
=>\(-\frac16<-\frac18<-\frac{1}{10}\)
=>\(-\frac16+1<-\frac18+1<-\frac{1}{10}+1\)
=>\(\frac56<\frac78<\frac{9}{10}\) <1
mà \(1=\frac88;1=\frac55<\frac65\)
nên \(\frac56<\frac78<\frac{9}{10}<\frac88<\frac65\)
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)
b)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)
\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)
a )
2 / 3 + 5 / 4 = 8 / 12 + 15 / 12 = 23 / 12
b )
15 / 25 + 6 / 21 = 3 / 5 + 2 / 7
= 21 / 35 + 10 / 35
= 31 / 35
c )
4 / 3 - 11 / 15 = 20 / 15 - 11 / 15
= 9 / 15
d )
12 / 10 - 2 / 5 = 12 / 10 - 4 / 10
= 8 / 10
= 4 / 5
1) \(\frac{4}{5}.\frac{8}{3}.\frac{x}{7}=\frac{96}{105}\)
\(\frac{32.x}{105}=\frac{96}{105}\)
\(32x=96\)
\(x=3\)
Bài 1: Tìm \(x\)
\(\frac45\times\frac83\times\) \(\frac{x}{7}\) = \(\frac{96}{105}\)
\(\frac{32}{15}\) x \(\frac{x}{7}\) = \(\frac{32}{35}\)
\(\frac{x}{7}\) = \(\frac{32}{35}\) : \(\frac{32}{15}\)
\(\frac{x}{7}\) = \(\frac{32}{35}\) x \(\frac{15}{32}\)
\(\frac{x}{7}\) = \(\frac37\)
\(x=\frac37\times7\)
\(x=3\)
b; \(\frac74\times\frac{3}{x}\times\) \(\frac{11}{5}\) = \(\frac{231}{200}\)
\(\frac74\) x \(\frac{3}{x}\) = \(\frac{231}{200}\) : \(\frac{11}{5}\)
\(\frac74\times\frac{3}{x}\) = \(\frac{231}{200}\) x \(\frac{5}{11}\)
\(\frac74\) x \(\frac{3}{x}\) = \(\frac{21}{40}\)
\(\frac{3}{x}\) = \(\frac{21}{40}\) : \(\frac74\)
\(\frac{3}{x}\) = \(\frac{21}{40}\) x \(\frac47\)
\(\frac{3}{x}\) = \(\frac{3}{10}\)
\(x=3:\frac{3}{10}\)
\(x=3\times\frac{3}{10}\)
\(x=10\)
a)\(\dfrac{4}{1}+\dfrac{5}{9}=\dfrac{36}{9}+\dfrac{5}{9}=\dfrac{41}{9}\)
b) \(\dfrac{8}{3}-\dfrac{2}{1}=\dfrac{8}{3}-\dfrac{6}{3}=\dfrac{2}{3}\)
\(\dfrac{5}{12}x\dfrac{4}{15}=\dfrac{20}{180}=\dfrac{1}{9}\)
\(\dfrac{5x8}{15}=\dfrac{40}{15}=\dfrac{8}{3}\)