1/2-(-1/3)+1/23+1/6
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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
a) A = 3² . 1/243 . 81² . 1/3²
= 3² . 1/3⁵ . (3⁴)² . 1/3²
= 3² . 1/3⁷ . 3⁸
= 3¹⁰ . 1/3⁷
= 3³
= 27
b) B = (4.2⁵) : (2³ . 1/6)
= (4.32) : (8 . 1/6)
= 128 : 4/3
= 96
c) C = (-1/3)³.(-1/3)².(-1/3)
= (-1/3)³⁺²⁺¹
= (-1/3)⁶
= 1/729
d) D = (-1/3)⁻¹ - (-6/7)⁰ + (1/2)² : 2
= -3 - 1 + 1/4 : 2
= -4 + 1/8
= -31/8
\(\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{23}}\times\dfrac{\dfrac{1}{3}-0,25-0,2}{1\dfrac{1}{6}-0,875-0,7}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{23}-\dfrac{1}{23}}\times\dfrac{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}-\dfrac{7}{10}}\)
\(=\dfrac{\dfrac{1}{3} +\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}\times2+\dfrac{1}{7}\times2-\dfrac{1}{23}\times2}\times\dfrac{\dfrac{2}{6}-\dfrac{2}{8}-\dfrac{2}{10}}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{2\times\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}\right)}\times\dfrac{2\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{1}{2}\times\dfrac{2}{7}\)
\(=\dfrac{1}{7}\)
Ta có: \(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
=0
\(\left(\dfrac{1}{57}-\dfrac{1}{5757}+\dfrac{1}{23}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)= 0
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
a: \(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}=\dfrac{15}{26}-\dfrac{4}{26}=\dfrac{11}{26}\)
b: \(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-5=\dfrac{-4}{7}-5+\dfrac{2}{9}=-\dfrac{337}{63}\)
c: \(=-\dfrac{11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
\(\dfrac{1}{2}-\left(-\dfrac{1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{23}+\dfrac{1}{6}=\dfrac{5}{6}+\dfrac{1}{6}+\dfrac{1}{23}=1+\dfrac{1}{23}=\dfrac{24}{23}\)
\(\dfrac{1}{2}-\left(\dfrac{-1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}\\ =\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}+\dfrac{1}{23}\\ =\left(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}\right)+\dfrac{1}{23}\\ =1+\dfrac{1}{23}\\ =\dfrac{24}{23}\)