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a) x + 2/5 = -4/3
x = -4/3 - 2/5
x = -26/15
b) -5/6 + 1/3 x = (-1/2)²
-5/6 + 1/3 x = 1/4
1/3 x = 1/4 + 5/6
1/3 x = 13/12
x = 13/12 : 1/3
x = 13/4
c) 7/12 - (x + 7/6) . 6/5 = (-1/2)³
7/12 - (x + 7/6) . 6/5 = -1/8
(x + 7/6) . 6/5 = 7/12 + 1/8
(x + 7/6) . 6/5 = 17/24
x + 7/6 = 17/24 : 6/5
x + 7/6 = 85/144
x = 85/144 - 7/6
x = -83/144
\(a,x+\dfrac{2}{5}=-\dfrac{4}{3}\\ \Rightarrow x=-\dfrac{26}{15}\\ b,-\dfrac{5}{6}+\dfrac{1}{3}x=\left(-\dfrac{1}{2}\right)^2\\ \Rightarrow-\dfrac{5}{6}+\dfrac{1}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{3}x=\dfrac{13}{12}\\ \Rightarrow x=\dfrac{13}{4}\\ c,\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\left(-\dfrac{1}{2}\right)^3\\ \Rightarrow\dfrac{7}{12}-\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=-\dfrac{1}{8}\\ \Rightarrow\left(x+\dfrac{7}{6}\right).\dfrac{6}{5}=\dfrac{17}{24}\\ \Rightarrow x+\dfrac{7}{6}=\dfrac{85}{144}\\ \Rightarrow x=-\dfrac{83}{144}.\)
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
a, \(\left(\dfrac{1}{3}\right)^2.9.3-\dfrac{1}{2}=\dfrac{1}{9}.9.3-\dfrac{1}{2}=3-\dfrac{1}{2}=\dfrac{5}{2}\)
b, \(\dfrac{1}{4}+\dfrac{3}{4}.\dfrac{2}{5}=\dfrac{1}{4}\left(1+\dfrac{6}{5}\right)=\dfrac{1}{4}.\dfrac{11}{5}=\dfrac{11}{20}\)
a) \(\dfrac{4}{9}+\dfrac{1}{4}=\dfrac{25}{36}\)
b) \(\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}\right)+\dfrac{1}{3}\cdot\left(-\dfrac{1}{5}\right)=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)=\dfrac{1}{3}\cdot-1=-\dfrac{1}{3}\)
c) \(\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(1-\dfrac{1}{2}\right)^2\right]=\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{5}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{1}{5}-0=\dfrac{1}{5}\)
`#3107.101107`
`a)`
\(\dfrac{4}{9}+\dfrac{1}{4}=\dfrac{16}{36}+\dfrac{9}{36}=\dfrac{25}{36}\)
`b)`
\(\dfrac{1}{3}\cdot\left(\dfrac{-4}{5}\right)+\dfrac{1}{3}\cdot\left(-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{3}\cdot\left(-1\right)\)
\(=-\dfrac{1}{3}\)
`c)`
\(\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(1-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{1}{5}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=\dfrac{1}{5}-0\)
\(=\dfrac{1}{5}\)