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TN
1
TL
5
17 tháng 8 2018
\(\frac{2}{5}+1=\frac{2}{5}+\frac{5}{5}=\frac{7}{5}\) ;\(\frac{6}{7}.4=\frac{24}{7}\);\(\frac{3}{7}:2=\frac{3}{7}.\frac{1}{2}=\frac{3}{14}\);\(2:\frac{2}{4}=2.\frac{4}{2}=2.2=4\);\(\frac{3}{4}.4+\frac{1}{3}=\frac{12}{4}+\frac{1}{3}=3+\frac{1}{3}=\frac{10}{3}\);\(\frac{1}{4}:\frac{1}{3}-\frac{1}{2}=\frac{1}{4}.\frac{3}{1}-\frac{1}{2}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)
CM
17 tháng 8 2018
\(\frac{2}{5}+1=\frac{2}{5}+\frac{5}{5}=\frac{7}{5}\)
\(\frac{6}{7}\times4=\frac{6\times4}{7}=\frac{24}{7}\)
\(\frac{3}{7}\div2=\frac{3}{7}\times\frac{1}{2}=\frac{3\times1}{7\times2}=\frac{3}{14}\)
\(2\div\frac{2}{4}=2\times\frac{4}{2}=2\times2=4\)
\(\frac{3}{4}\times4+\frac{1}{3}=\frac{3\times4}{4}+\frac{1}{3}=3+\frac{1}{3}=\frac{9}{3}+\frac{1}{3}=\frac{10}{3}\)
\(\frac{1}{4}\div\frac{1}{3}-\frac{1}{2}=\frac{1}{4}\times3-\frac{1}{2}=\frac{1\times3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\)
F
23 tháng 1 2020
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
M
23 tháng 1 2020
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
25 tháng 2 2020
1/2 + 4/3 = 3/6 + 8/6 = 11/6
4/5 + 3/4 : 2/3 = 4/5 + 3/4 * 3/2 = 4/5 + 9/8 = 32/40 + 45/40 = 77/40
7/2 - 4/3 * 1/6 = 7/2 - 2/9 = 63/18 - 4/18 = 59/18
4/5 * 4/7 : 2/3 = 16/35 : 2/3 = 16/35 * 3/2 = 24/35
1/4 : 5/3 + 1/6 = 1/4 * 3/5 + 1/6 = 3/20 + 1/6 = 9/60 + 10/60 = 19/60
3 - 4/5 : 1/3 = 3 - 4/5 * 3/1 = 3 - 12/5 = 15/5 - 12/5 = 3/5
5/9 * 2/7 - 3/7 = 10/63 - 3/7 = 10/63 - 27/63 = -17/63
TT
0
7 tháng 6 2016
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\)= \(\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}\)= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\)= \(\frac{1.2.3}{2.3.4}=\frac{1}{4}\)
\(\left(1+\frac{1}{2}\right):\left(1+\frac{1}{3}\right):\left(1+\frac{1}{4}\right)\)= \(\frac{2+1}{2}:\frac{3+1}{3}:\frac{4+1}{4}\)= \(\frac{3}{2}:\frac{4}{3}:\frac{5}{4}=\frac{3}{2}.\frac{3}{4}.\frac{4}{5}=\frac{3.3.4}{2.4.5}=\frac{3.3}{2.5}=\frac{9}{10}\)
\(\frac{6}{5}-\frac{4}{5}.\frac{4}{8}+1=\frac{6}{5}+1-\frac{4.4}{5.8}=\frac{6+5}{5}-\frac{4}{5.2}=\frac{11}{5}-\frac{2}{5}=\frac{9}{5}\)
x < \(\frac{3}{4}+\frac{4}{5}=\frac{15}{20}+\frac{16}{20}=\frac{31}{20}=\frac{20}{20}+\frac{11}{20}=1\frac{11}{20}\)=> x = 0 ; 1
\(\frac{1}{2}+\frac{2}{3}< x< 4-\frac{5}{4}\)<=> \(\frac{3}{6}+\frac{4}{6}< x< \frac{16}{4}-\frac{5}{4}\)<=> \(\frac{7}{6}< x< \frac{11}{4}\)=> \(\frac{14}{12}< \frac{12x}{12}< \frac{33}{12}\)
=> 14 < 12x < 33 => 12x = 24 => x = 24 : 12 = 2
4/1-4/3= 12/3 - 4/3= 8/3
\(\dfrac{4}{1}-\dfrac{4}{3}\\ =\dfrac{4\times3}{1\times3}-\dfrac{4}{3}\\ =\dfrac{12}{3}-\dfrac{4}{3}\\ =\dfrac{12-4}{3}\\ =\dfrac{8}{3}\)