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`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
a: =9+7=16
b: =11+3/13-2-4/7-5-3/13
=4-4/7
=28/7-4/7=24/7
c: =2/7(5+1/4-3-1/4)=2/7x2=4/7
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: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
3) \(58\left(3\dfrac{1}{29}-2\dfrac{1}{58}\right):\dfrac{1}{3}\)
\(=58\left(\dfrac{88}{29}-\dfrac{117}{58}\right)\cdot3\\ =\left(58\cdot\dfrac{88}{29}-58\cdot\dfrac{117}{58}\right)\cdot3\)
\(=\left(176-117\right)\cdot3\\ =59\cdot3\\ =177\)
4) \(A=1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\cdot1\dfrac{1}{5}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}\\ =\dfrac{3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5}\\ =\dfrac{6}{2}\\ =3\)
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
Sửa đề: \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\)
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+.....+\dfrac{1}{2^{58}}\)
\(\Rightarrow2^3A=2^3.\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+\dfrac{1}{2^{58}}\right)\)
\(\Rightarrow2^3A=1+\dfrac{1}{2}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{55}}\)
\(\Rightarrow2^3A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\right)\)\(\Rightarrow7A=1-\dfrac{1}{2^{58}}\Rightarrow A=\dfrac{1-\dfrac{1}{2^{58}}}{7}\)
Vậy...........
~ Học tốt nha ~
\(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\\ 2^3\cdot A=\dfrac{2^3}{2}+\dfrac{2^3}{2^4}+\dfrac{2^3}{2^7}+...+\dfrac{2^3}{2^{58}}\\ 8A=4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\\ 8A-A=\left(4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\right)\\ 7A=4-\dfrac{1}{2^{58}}\\ A=\dfrac{4-\dfrac{1}{2^{58}}}{7}\)