\(2+4:\left(\dfrac{2}{3}-\dfrac{1}{6}\right).\left(-2,25\right)\)
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\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)
\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)
\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
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Giải:
a) \(\left(\dfrac{1}{3}.x\right):\dfrac{2}{3}=1\dfrac{3}{4}:\dfrac{2}{5}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{35}{8}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{8}.\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{12}\)
\(\Leftrightarrow x=\dfrac{35}{12}:\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{35}{4}\)
Vậy \(x=\dfrac{35}{4}\).
b) \(4,5:0,3=2,25\left(0,1.x\right)\)
\(\Leftrightarrow15=2,25\left(0,1.x\right)\)
\(\Leftrightarrow2,25\left(0,1.x\right)=15\)
\(\Leftrightarrow0,1.x=\dfrac{15}{2,25}\)
\(\Leftrightarrow0,1.x=\dfrac{20}{3}\)
\(\Leftrightarrow x=\dfrac{20}{3}:0,1\)
\(\Leftrightarrow x=\dfrac{200}{3}\)
Vậy \(x=\dfrac{200}{3}\).
c) \(8:\left(\dfrac{1}{4}.x\right)=2:0,02\)
\(\Leftrightarrow8:\left(\dfrac{1}{4}.x\right)=100\)
\(\Leftrightarrow\dfrac{1}{4}.x=\dfrac{2}{25}\)
\(\Leftrightarrow x=\dfrac{2}{25}:\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Vậy \(x=\dfrac{8}{25}\).
d) \(3:2\dfrac{1}{4}=\dfrac{3}{4}:\left(6.x\right)\)
\(\Leftrightarrow3:\dfrac{9}{4}=\dfrac{3}{4}:\left(6.x\right)\)
\(\Leftrightarrow\dfrac{4}{3}=\dfrac{3}{4}:\left(6.x\right)\)
\(\Leftrightarrow\dfrac{4}{3}:\left(6.x\right)=\dfrac{3}{4}\)
\(\Leftrightarrow6.x=\dfrac{4}{3}:\dfrac{3}{4}\)
\(\Leftrightarrow6.x=\dfrac{16}{9}\)
\(\Leftrightarrow x=\dfrac{16}{9}:6\)
\(\Leftrightarrow x=\dfrac{8}{27}\)
Vậy \(x=\dfrac{8}{27}\).
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a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)
Vậy: \(x\in\varnothing\)
a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)
hay x=1
Vậy: x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1