tính 20% + \(\dfrac{2}{3}\) +1.8+ 2\(\dfrac{1}{3}\)
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Những câu hỏi liên quan
1 tháng 3 2018
Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
27 tháng 6 2023
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
O
5 tháng 3 2022
\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)
\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)
\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)
\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)
5 tháng 3 2022
\(\dfrac{-1}{6}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-11}{15}\)
\(\dfrac{-9}{8}\)
HN
1
16 tháng 4 2022
\(\dfrac{3}{7}.\left(-\dfrac{2}{5}\right).2\dfrac{1}{3}.20.\dfrac{19}{72}\)
\(=\dfrac{3}{7}.\left(-\dfrac{2}{5}\right).\dfrac{7}{3}.20.\dfrac{19}{72}\)
\(=\left(-\dfrac{6}{35}\right).\dfrac{7}{3}.20.\dfrac{19}{72}\)
\(=\left(-\dfrac{2}{5}\right).20.\dfrac{19}{72}\)
\(=\left(-8\right).\dfrac{19}{72}\)
\(=-\dfrac{19}{9}\)
TM
1
26 tháng 3 2022
\(1=2-1=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)=>\(\dfrac{1}{1+\sqrt{2}=}=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}-1\)
cmtt thì biểu thức thành
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{21}-\sqrt{20}\)=\(1+\sqrt{21}\)
GL
24 tháng 8 2023
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\dfrac{3}{12}+\dfrac{1}{4}=\dfrac{3:3}{12:3}+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\dfrac{4}{10}+\dfrac{3}{5}=\dfrac{4:2}{10:2}+\dfrac{3}{5}=\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\)
\(\dfrac{12}{27}+\dfrac{2}{9}=\dfrac{12:3}{27:3}+\dfrac{2}{9}=\dfrac{4}{9}+\dfrac{2}{9}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{7}{3}+\dfrac{20}{15}=\dfrac{7}{3}+\dfrac{20:5}{15:5}=\dfrac{7}{3}+\dfrac{4}{3}=\dfrac{11}{3}\)
\(20\%+\dfrac{2}{3}+1,8+2\dfrac{1}{3}\\ =\dfrac{1}{5}+\dfrac{2}{3}+\dfrac{9}{5}+\dfrac{7}{3}\\ =\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{2}{3}+\dfrac{7}{3}\right)\\ =\dfrac{10}{5}+\dfrac{9}{3}\\ =2+3\\ =5\)