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S
20 tháng 12 2022
g)\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\left(-\dfrac{3}{4}+-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right).\dfrac{7}{3}\)
\(=\left(-1+1\right).\dfrac{7}{3}=0.\dfrac{7}{3}=0\)
KV
20 tháng 12 2022
f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)
\(=\dfrac{3^3.5^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)
\(=\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)
\(=\dfrac{5^3}{6^3}\)
\(=\dfrac{125}{216}\)
VN
20 tháng 12 2022
\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=-1+1-\dfrac{1}{3}\)
\(=0-\dfrac{1}{3}\)
\(=\dfrac{-1}{3}\)
------------------------------------------
\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)
\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)
\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)
\(=21+\left(-3\right)\)
\(=18\)
------------------------------------------------
\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)
\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)
\(=\sqrt{\dfrac{29}{36}}\)
\(=\dfrac{\sqrt{29}}{6}\)
S
20 tháng 12 2022
\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)
22 tháng 12 2021
\(a,=\dfrac{9}{12}-\dfrac{7}{12}-\dfrac{8}{12}=-\dfrac{1}{2}\\ b,=\dfrac{95}{6}+\dfrac{17}{18}-\dfrac{5}{12}=\dfrac{589}{36}\)
LV
3
12 tháng 12 2021
\(\dfrac{5}{6}.18\dfrac{2}{3}-\dfrac{5}{6}.6\dfrac{2}{3}=\dfrac{5}{6}\left(18\dfrac{2}{3}-6\dfrac{2}{3}\right)=\dfrac{5}{6}.12=10\)
12 tháng 12 2021
\(=\left(18\dfrac{2}{3}-6\dfrac{2}{3}\right).\dfrac{5}{6}=\dfrac{5}{6}.12=10\)
NT
2
11 tháng 12 2017
\(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)
= \(\dfrac{\left(5.3\right)^3+5.\left(5.3\right)^2-5^3}{\left(6.3\right)^3+6.\left(6.3\right)^2-6^3}\)
= \(\dfrac{5^3.3^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)
= \(\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)
=\(\dfrac{5^3}{6^3}=\dfrac{125}{216}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 7 2023
a) \(\dfrac{27}{2}\cdot7,5+\dfrac{27}{2}\cdot2,5-150\)
\(=\dfrac{27}{2}\cdot\left(7,5+2,5\right)-150\)
\(=\dfrac{27}{2}\cdot10-150\)
\(=135-150\)
\(=-15\)
b) \(3^3\cdot\dfrac{18}{5}-3^3\cdot2\dfrac{2}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\dfrac{18}{5}-3^3\cdot\dfrac{12}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{12}{5}-\dfrac{6}{5}\right)\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{18}{5}\right)\)
\(=3^3\cdot0\)
\(=0\)