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Bài 1:
a) \(2\frac{1}{4}+1\frac{1}{2}-1\frac{4}{5}\)
=\(\frac{9}{4}\)+\(\frac{3}{2}\)-\(\frac{9}{5}\)
=\(\frac{45}{20}+\frac{30}{20}-\frac{36}{20}\)
=\(\frac{39}{20}\)
b) \(2\frac{2}{3}\times3\frac{1}{4}:2\frac{3}{4}\)
\(\frac{8}{3}\times\frac{13}{4}:\frac{11}{4}\)
=\(\frac{8}{3}\times\frac{13}{4}\times\frac{4}{11}\)
=\(\frac{104}{33}\)
Bài 2:
a) \(\frac{7}{9}+\frac{4}{5}+\frac{11}{9}+\frac{6}{5}\)
=\(\left(\frac{7}{9}+\frac{11}{9}\right)+\left(\frac{4}{5}+\frac{6}{5}\right)\)
=2+2
=4
b) \(\frac{21}{36}\times\frac{39}{14}\times\frac{54}{13}\)
=\(\frac{21\times39\times54}{36\times14\times13}\)
=\(\frac{675}{100}\)
\(3\frac{8}{9}=\frac{3\cdot9+8}{9}=\frac{27+8}{9}=\frac{35}{9}\)
Ok nhé bạn
Bài 1;
\(\frac{5}{7}=\frac{10}{14}\); \(\frac{6}{7}=\frac{12}{14}\)
=> \(\frac{10}{14}< \frac{11}{14}< \frac{12}{14}=>\frac{5}{7}< \frac{11}{14}< \frac{6}{7}\)
2 và 4 phần 5 : 14/5
5 và 1 phần 4 : 21/4
7 và 2 phần 5 : 37/5
4 và 1 phần 25 : 101/25
\(2\frac{4}{5}=2,8\), \(5\frac{1}{4}=5,25\),\(7\frac{2}{5}=7,4\),\(4\frac{1}{25}=4,04\)
\(5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\times4\dfrac{1}{2}-2\times2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}\times\dfrac{9}{2}-2\times\dfrac{7}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left[\dfrac{7}{3}\times\left(\dfrac{9}{2}-2\right)\right]:\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\left(\dfrac{7}{3}\times\dfrac{5}{2}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}\cdot\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{59}{15}-\dfrac{50}{15}\)
\(=\dfrac{9}{15}\)
\(=\dfrac{3}{5}\)
\(Toru\)
\(\frac{4}{9}:8-\left(\frac{1}{4}+\frac{9}{20}\right):1\frac{4}{5}\)
\(=\frac{1}{18}-\frac{7}{10}\cdot\frac{5}{9}=\frac{1}{18}-\frac{7}{18}=-\frac{1}{3}\)
\(\frac{4}{9}:8-\left(\frac{1}{4}+\frac{9}{20}\right):1\frac{4}{5}\)
\(=\frac{1}{18}-\frac{5+9}{20}\times\frac{5}{9}\)
\(=\frac{1}{18}-\frac{14\times5}{20\times9}\)
\(\frac{1}{18}-\frac{14}{36}=\frac{2-14}{36}=\frac{-12}{36}=\frac{-1}{3}\)