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\(\left|x-\dfrac{1}{2}\right|-1=\dfrac{5}{2}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{5}{2}+1\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{7}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{7}{2}\\x-\dfrac{1}{2}=-\dfrac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(9\times\left(\dfrac{-1}{3}\right)^3+\dfrac{1}{3}:0,25-5\dfrac{3}{4}=\dfrac{9}{1}\times\left(\dfrac{-1}{27}\right)+\dfrac{1}{3}:\dfrac{1}{4}-\dfrac{23}{4}\)
\(=\left(\dfrac{-9}{27}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\left(\dfrac{-1}{3}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\dfrac{3}{3}-\dfrac{23}{4}=1-\dfrac{23}{4}=\dfrac{4}{4}-\dfrac{23}{4}=\dfrac{-19}{4}\)
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)\(\left(-2\right)^3\cdot\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
=\(-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
=\(-8\cdot\dfrac{1}{2}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)\)
=\(-4:\dfrac{13}{12}\)
=\(-4\cdot\dfrac{12}{13}\)
=\(-\dfrac{48}{13}\)
\(\text{A=}\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)
\(A=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)
\(A=\left(\dfrac{7}{8}+\dfrac{-1}{4}\right):\dfrac{1}{144}\)
\(A=\dfrac{5}{8}.144=90\)
\(\text{B=}\dfrac{3}{4}.1\dfrac{4}{9}-\left(\dfrac{-3}{4}\right)\)
\(B=\dfrac{3}{4}.\dfrac{13}{9}+\dfrac{3}{4}\)
\(B=\dfrac{13}{12}+\dfrac{3}{4}\)
\(B=\dfrac{11}{6}\)
a: \(=\dfrac{9}{13}\cdot\dfrac{4}{5}=\dfrac{36}{65}\)
b: \(=\dfrac{-7}{10}:\dfrac{3}{2}=\dfrac{-7}{10}\cdot\dfrac{2}{3}=\dfrac{-14}{30}=-\dfrac{7}{15}\)
c: \(=\dfrac{7}{6}\left(3+\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{7}{6}\cdot3=\dfrac{7}{2}\)
\(D=\left(-8\right).\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=\left(-8\right)\dfrac{1}{2}:\dfrac{13}{12}=\left(-4\right).\dfrac{12}{13}=-\dfrac{48}{13}\)
a, \(\dfrac{3}{4}x=-\dfrac{9}{8}\)
x= \(-\dfrac{3}{2}\)
b, |x| + 0,25= 5,25
|x | = 5
=> x\(\in\){ +- 5}
Ko chắc đúng, kiểm tra trc khi làm
\(\dfrac{2x-1}{3}=\dfrac{-5}{0.6}\)
\(\Leftrightarrow2x-1=-25\)
hay x=-12
Ta lần lượt có:
\(\dfrac{1}{0,25}=\dfrac{100}{25}=4;\) \(\left(1\dfrac{1}{4}\right)^2=\left(\dfrac{5}{4}\right)^2=\dfrac{25}{16};\)
\(\dfrac{1}{\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{\dfrac{16}{9}}=\dfrac{9}{16};\) \(\left(\dfrac{5}{4}\right)^3=\dfrac{125}{64}\) và \(\dfrac{1}{\left(-\dfrac{2}{3}\right)^3}=\dfrac{1}{-\dfrac{8}{27}}=-\dfrac{27}{8}\)
Vậy \(P=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\left(-\dfrac{27}{8}\right)\)
\(=\dfrac{25}{4}-25.\dfrac{9}{16}.\dfrac{64}{125}.\dfrac{8}{27}\)
\(=\dfrac{25}{4}-\dfrac{32}{15}\)
\(=\dfrac{375-128}{60}=\dfrac{247}{60}\)
P=4.(25/16)+25.(9/16:25/16):(-27/8)
=25/4 +25.(9/25):(-27/8)
=25/4 +9:(-27/8)
=25/4 +(-8/3)
=43/12
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=1\\x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left|x-\dfrac{1}{3}\right|-0,25=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|-\dfrac{1}{4}=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}+\dfrac{1}{4}=1\)
\(\text{Vậy }x-\dfrac{1}{3}=1\)
\(x\) \(=1+\dfrac{1}{3}=\dfrac{4}{3}\)
\(\text{hoặc }x-\dfrac{1}{3}=-1\)
\(x\) \(=\left(-1\right)+\dfrac{1}{3}=\dfrac{-2}{3}\)
\(\Rightarrow x\in\left\{\dfrac{4}{3};\left(\dfrac{-2}{3}\right)\right\}\)