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a; \(\dfrac{1}{2}-\dfrac{-3}{6}+\dfrac{5}{3}-\dfrac{9}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{5}{3}-\dfrac{3}{4}\)
\(=1-\dfrac{3}{4}+\dfrac{5}{3}=\dfrac{1}{4}+\dfrac{5}{3}=\dfrac{3+20}{12}=\dfrac{23}{12}\)
b: \(=\dfrac{3}{11}\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}\cdot\dfrac{-6-16}{9}=\dfrac{3}{11}\cdot\dfrac{-22}{9}=\dfrac{-2}{3}\)
c: \(=1-3+\dfrac{1}{4}=-2+\dfrac{1}{4}=-\dfrac{7}{4}\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
a) \(0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\left(-\dfrac{4}{5}\right)\)
\(\Rightarrow\dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{2}{3}=x-\dfrac{4}{5}\)
\(\Leftrightarrow15x-20=30x-24\)
\(\Leftrightarrow15x-30x=-24+20\)
\(\Leftrightarrow-15x=-4\)
\(\Rightarrow x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{4}{15}\)
b) \(\dfrac{2}{3}\cdot\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{5}{18}=\dfrac{2}{3}-x\)
\(\Leftrightarrow6x+5=12-18x\)
\(\Leftrightarrow6x+18x=12-5\)
\(\Leftrightarrow24x=7\)
\(\Rightarrow x=\dfrac{7}{24}\)
Vậy \(x=\dfrac{7}{24}\)
\(a,0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-x=\dfrac{-4}{5}+\dfrac{1}{6}+\dfrac{1}{2}\\ x\left(\dfrac{1}{2}-1\right)=\dfrac{-24}{30}+\dfrac{5}{30}+\dfrac{15}{30}\\ \dfrac{-1}{2}x=\dfrac{-2}{15}\\ x=\dfrac{-2}{15}:\dfrac{-1}{2}\\ x=\dfrac{-2}{15}\cdot\left(-2\right)\\ x=\dfrac{4}{15}\)
\(b,\dfrac{2}{3}\cdot\left(\dfrac{1}{2}\cdot x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x+x=\dfrac{2}{3}+\dfrac{2}{9}-\dfrac{1}{2}\\ x\left(\dfrac{1}{3}+1\right)=\dfrac{12}{18}+\dfrac{4}{18}-\dfrac{9}{18}\\ \dfrac{4}{3}x=\dfrac{7}{18}\\ x=\dfrac{7}{18}:\dfrac{4}{3}\\ x=\dfrac{7}{18}\cdot\dfrac{3}{4}\\ x=\dfrac{7}{24}\)
a) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)
\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)
\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)
x = \(\dfrac{1}{10}\)-\(\dfrac{7}{20}\)
x = \(\dfrac{-1}{4}\)
Chúc bn học tốt
a)
\(\left[\dfrac{-2}{3}+0,5:\left(\dfrac{-3}{2}\right)^2\right]+\left[1\dfrac{1}{5}-1,4.\dfrac{5}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{1}{2}:\dfrac{9}{4}\right]+\left[\dfrac{6}{5}-\dfrac{7}{5}.\dfrac{5}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{1}{2}.\dfrac{4}{9}\right]+\left[\dfrac{6}{5}-\dfrac{7}{6}+6\right]\\ =\left[\dfrac{-2}{3}+\dfrac{2}{3}\right]+\left[\dfrac{36}{30}-\dfrac{35}{30}+6\right]\\ =0+\left[\dfrac{1}{30}+6\right]\\ =6\dfrac{1}{30}\)
b)
\(\left(-0,2\right)^2.5-\dfrac{8^2.9^5}{3^9.4^3}\\ =0,4.5-\dfrac{\left(2^3\right)^2.\left(3^2\right)^5}{3^9.\left(2^2\right)^3}\\ =2-\dfrac{2^6.3^{10}}{3^9.2^6}\\ =2-\dfrac{1.3}{1.1}\\ =2-3\\ =-1\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
1.
\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{1}{6}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{1}{9}\right]\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{9}\right\}\)
=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{10}{27}\right\}\)
=\(\dfrac{2}{3}.\dfrac{8}{27}\)
=...