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11 tháng 7 2017
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
11 tháng 7 2017
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
GM
6
30 tháng 4 2019
\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)
\(=\frac{-48}{13}\)
T
1 tháng 10 2023
a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)
\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)
\(=\dfrac{10}{12}\)
\(=\dfrac{5}{6}\)
\(---\)
b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)
\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)
\(=18+\dfrac{3015}{2}\)
\(=\dfrac{36}{2}+\dfrac{3015}{2}\)
\(=\dfrac{3051}{2}\)
\(---\)
c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)
\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)
\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\dfrac{4}{6}-\dfrac{1}{6}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\)
\(---\)
d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)
\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)
\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)
\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)
\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)
\(=-\dfrac{25}{41}\)
#\(Toru\)
5 tháng 3 2018
\(x-0.25=-\frac{6}{11}\cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-\frac{1}{4}=-\frac{6}{11}\cdot\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right)\)
\(x-\frac{1}{4}=-\frac{6}{11}\cdot\frac{11}{12}\)
\(x-\frac{1}{4}=-\frac{1}{2}\)
\(x=-\frac{1}{2}+\frac{1}{4}\)
\(x=-\frac{2}{4}+\frac{1}{4}\)
\(x=-\frac{1}{4}\)
Vậy .......
5 tháng 3 2018
\(x-0,25=-\frac{6}{11}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-0,25=-\frac{6}{11}.\frac{11}{12}\)
\(x-0,25=-\frac{1}{2}\)
\(x=-\frac{1}{2}+0,25=-\frac{1}{4}\)
học tốt ~~~
SB
1
NH
9 tháng 7 2023
`5/6+6/7-1/6+7/3`
`=(5/6-1/6+7/3)+6/7`
`=(4/6+7/3)+6/7`
`=(2/3+7/3)+6/7`
`=9/3+6/7`
`=3+6/7`
`=21/7+6/7`
`=27/7`
\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(\frac{2}{\frac{1}{4}}-\frac{1}{\frac{1}{6}}\right)\)
\(=-2^3.\frac{\frac{3}{4}-0,25}{\frac{2}{\frac{1}{4}}-\frac{1}{\frac{1}{6}}}\)
\(=-2^3.\frac{\frac{3}{4}-0,25}{2}\)
\(=-\frac{\left(\frac{3}{4}-0,25\right).2^3}{2}\)
\(=-2^2.\left(\frac{3}{4}-0,25\right)\)
= -4.0,5
= -2