(1,5 điểm) Tính.
$a)$ $\dfrac{4}{9}+\dfrac{1}{4}$.
$b)$ $\dfrac{1}{3}.\Big( \dfrac{-4}{5} \Big) + \dfrac{1}{3}.\dfrac{-1}{5}$.
$c)$ $\dfrac{1}{5}-\Big[\dfrac{1}{4}-\Big(1-\dfrac{1}{2}\Big)^2\Big]$.
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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) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
`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:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
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)`
\(\dfrac{1}{5}-\left[\dfrac{1}{4}-\left(1-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{1}{5}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=\dfrac{1}{5}-0\)
\(=\dfrac{1}{5}\)