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17 tháng 6 2016

\(A=\frac{1,6:\left(1+2\right)}{3}\)

\(A=\frac{1,6:3}{3}\)

\(A=\frac{8}{45}\)

17 tháng 6 2016

1,6 :  3= 8/15 k cái ủng hộ

=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023

=0+0+...+0-1-2023

=-2024

6 tháng 3

Bằng 2024

 

a: \(2\dfrac{3}{5}+1\dfrac{2}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{26}{10}+\dfrac{217}{10}=\dfrac{243}{10}\)

b: \(4\dfrac{3}{4}-3\dfrac{2}{3}:1\dfrac{1}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}\cdot\dfrac{6}{7}\)

\(=\dfrac{19}{4}-\dfrac{22}{7}\)

\(=\dfrac{19\cdot7-22\cdot4}{28}=\dfrac{45}{28}\)

1 tháng 5 2022

A = ( 1/33 - 13/55 + 17/777) . 0

A = 0

a: A=x^2y(2/3+3+1)=14/3*x^2y

=14/3*3^2*(-1/7)

=-2*3=-6

NV
29 tháng 4 2021

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)

\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)

\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)

\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)

\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)

\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)

\(A=2-\)\(\dfrac{1}{2^{2012}}\)

18 tháng 3 2023

a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

\(=\dfrac{9}{5}+1\)

\(=\dfrac{14}{5}\)

b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

18 tháng 3 2023

\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

\(=\dfrac{9}{5}+1\)

\(=\dfrac{9}{5}+\dfrac{5}{5}\)

\(=\dfrac{14}{5}\)

\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)

13 tháng 10 2021

\(A=2\sin^2a+5\cos^2a=5\left(\sin^2a+\cos^2a\right)-3\sin^2a\\ A=5\cdot1-3\cdot\dfrac{4}{9}=5-\dfrac{4}{3}=\dfrac{11}{3}\)