\(\dfrac{1}{25}+\dfrac{2}{25}+\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{6}{25}+\dfrac{7}{25}+\dfrac{8}{25}+\dfrac{9}{25}+\dfrac{10}{25}\)
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GL
24 tháng 8 2023
a) \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{7}{9};\dfrac{5}{9}+\dfrac{2}{9}=\dfrac{7}{9}\)
Vậy: \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{5}{9}+\dfrac{2}{9}\)
b) \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{14}{25};\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}=\dfrac{14}{25}\)
Vậy: \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}\)
S
6 tháng 2 2023
a)\(\dfrac{7}{-25}+-\dfrac{8}{25}=-\dfrac{15}{25}=-\dfrac{3}{5}\)
b)\(\dfrac{7}{21}-\dfrac{9}{-36}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)
c)\(-\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(=-\dfrac{1}{2}+1\)
\(=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)
VT
6 tháng 2 2023
\(a,\dfrac{7}{-25}+\dfrac{-8}{25}=\dfrac{-7}{25}+\dfrac{-8}{25}=\dfrac{-15}{25}=\dfrac{-3}{5}\\ b,\dfrac{7}{21}-\dfrac{9}{-36}=\dfrac{7}{21}+\dfrac{9}{36}=\dfrac{7}{12}\\ c,\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\\ =-\dfrac{1}{2}+1\\ =\dfrac{1}{2}\)
8 tháng 7 2021
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
27 tháng 9 2021
\(=\dfrac{25}{9}.\dfrac{3}{10}+\dfrac{25}{9}.\dfrac{7}{10}+\dfrac{25}{3}=\dfrac{25}{9}\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\dfrac{25}{3}=\dfrac{25}{9}+\dfrac{25}{3}=\dfrac{100}{9}\)
BN
11 tháng 3 2022
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
27 tháng 5 2022
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
15 tháng 2 2021
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
=50/25=2
55/25 = 11/5