Tính:
1. 7 15 + 4 5 =
2. 4 5 × 5 8 =
3. 2 3 − 3 8 =
4. 1 2 : 2 5 =
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1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
\(\frac{6}{5}-\frac{1}{4}+\frac{4}{5}-\frac{3}{4}\)
\(=\left(\frac{6}{5}+\frac{4}{5}\right)-\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=2-1\)
\(=1\)
\(\frac{7}{9}-\frac{5}{12}+\frac{2}{9}-\frac{7}{12}\)
\(=\left(\frac{7}{9}+\frac{2}{9}\right)-\left(\frac{5}{12}+\frac{7}{12}\right)\)
\(=1-1\)
\(=0\)
các câu sau tương tự
A=1
B=0
C=ÂM 8/15
D=0
E=2
F=3/2
H=2/3
LẦN SAU CHO KHÓ SÍU NHA LỚP 5 CŨNG LÀM ĐC
a) 7/15 + 4/5 = 7/15 + 12/15 = 19/15
b) 4/5 x 5/8= 1/2
c) 2/3 - 3/8= 16/24 - 9/24 = 7/24
d) 1/2 : 2/5= 1/2 x 5/2= 5/4
a) \(\dfrac{7}{15}+\dfrac{4}{5}=\dfrac{7}{15}+\dfrac{12}{15}=\dfrac{19}{15}\)
b) \(\dfrac{4}{5}\times\dfrac{5}{8}=\dfrac{20}{40}=\dfrac{1}{2}\)
c) \(\dfrac{2}{3}-\dfrac{3}{8}=\dfrac{16}{24}-\dfrac{9}{24}=\dfrac{7}{24}\)
d) \(\dfrac{1}{2}:\dfrac{2}{5}=\dfrac{1}{2}\times\dfrac{5}{2}=\dfrac{5}{4}\)
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)
2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)
\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)
3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)
1. 19 15
2. 1 2
3. 7 24
4. 5 4