lx-\(\dfrac{1}{3}\)l - 0,25 = \(\dfrac{3}{4}\)
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\(\left|x-\dfrac{1}{2}\right|-1=\dfrac{5}{2}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{5}{2}+1\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{7}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{7}{2}\\x-\dfrac{1}{2}=-\dfrac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
a,(3/5+0,415-3/200).\(2\dfrac{2}{3}\).0,25
=(0,6+0.415-3/200)8/3.1/4
=(1,015-3/200).8/3.1/4
=(1015/1000-3/200).2/3
=(203/200-3/200).2/3
=1.2/3=2/3
b,0,25:(10,3-9,8)-3/4
=0,25:910,3-9,8)-0.75
=0,25:0,5-0,75
=-0,25
Mình làm từng đó trước,lần sau mình sẽ lm típ nha!
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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) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)
\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)
\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)
\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)
\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)
\(=-\dfrac{25}{41}\)
#\(Toru\)
\(9\times\left(\dfrac{-1}{3}\right)^3+\dfrac{1}{3}:0,25-5\dfrac{3}{4}=\dfrac{9}{1}\times\left(\dfrac{-1}{27}\right)+\dfrac{1}{3}:\dfrac{1}{4}-\dfrac{23}{4}\)
\(=\left(\dfrac{-9}{27}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\left(\dfrac{-1}{3}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\dfrac{3}{3}-\dfrac{23}{4}=1-\dfrac{23}{4}=\dfrac{4}{4}-\dfrac{23}{4}=\dfrac{-19}{4}\)
Thực hiện phép tính bằng cách tính hợp lí
a)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\)
=\(\dfrac{15}{34}+\dfrac{1}{3}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)
=\(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)
=\(1+1-\dfrac{32}{17}\)
=\(\dfrac{34}{17}-\dfrac{32}{17}\)
=\(\dfrac{2}{17}\)
b)\(\left(-2\right)^3\cdot\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
=\(-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
=\(-8\cdot\dfrac{1}{2}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)\)
=\(-4:\dfrac{13}{12}\)
=\(-4\cdot\dfrac{12}{13}\)
=\(-\dfrac{48}{13}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
\(\text{A=}\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)
\(A=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)
\(A=\left(\dfrac{7}{8}+\dfrac{-1}{4}\right):\dfrac{1}{144}\)
\(A=\dfrac{5}{8}.144=90\)
\(\text{B=}\dfrac{3}{4}.1\dfrac{4}{9}-\left(\dfrac{-3}{4}\right)\)
\(B=\dfrac{3}{4}.\dfrac{13}{9}+\dfrac{3}{4}\)
\(B=\dfrac{13}{12}+\dfrac{3}{4}\)
\(B=\dfrac{11}{6}\)
=\(16.\left(\dfrac{5}{8}-0,25\right):\left(3\dfrac{1}{3}-2\dfrac{1}{4}\right)\)
=\(16.\left(\dfrac{5}{8}-\dfrac{1}{4}\right):\left(\dfrac{10}{3}-\dfrac{9}{4}\right)\)
=\(16.\dfrac{3}{8}:\dfrac{13}{12}\)
=\(6:\dfrac{13}{12}\)
=\(\dfrac{72}{13}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=1\\x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left|x-\dfrac{1}{3}\right|-0,25=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|-\dfrac{1}{4}=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}+\dfrac{1}{4}=1\)
\(\text{Vậy }x-\dfrac{1}{3}=1\)
\(x\) \(=1+\dfrac{1}{3}=\dfrac{4}{3}\)
\(\text{hoặc }x-\dfrac{1}{3}=-1\)
\(x\) \(=\left(-1\right)+\dfrac{1}{3}=\dfrac{-2}{3}\)
\(\Rightarrow x\in\left\{\dfrac{4}{3};\left(\dfrac{-2}{3}\right)\right\}\)