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a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
\(\frac{4\frac{1}{2}.5\frac{2}{3}}{6\frac{3}{4}}=\frac{\frac{9}{2}.\frac{17}{3}}{\frac{27}{4}}=\frac{\frac{153}{6}}{\frac{27}{4}}=\frac{153}{6}.\frac{4}{27}=\frac{34}{9}\)
\(\frac{4\frac{1}{2}\times5\frac{2}{3}}{6\frac{3}{4}}\)
\(\frac{\frac{4\times2+1}{2}+\frac{5\times3+2}{3}}{\frac{6\times4+3}{4}}\)
\(=\frac{\frac{9}{2}+\frac{17}{3}}{\frac{27}{4}}\)
\(=\frac{\frac{27}{6}+\frac{34}{6}}{\frac{27}{4}}\)
\(=\frac{61}{6}\div\frac{27}{4}=\frac{61}{6}\times\frac{4}{27}=\frac{244}{162}=\frac{122}{81}\)
a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)
b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)
vậy x=25
1.
a) \(\frac{x}{4}=\frac{16}{x^2}\)
\(\Rightarrow x^3=64\)
\(\Rightarrow x^3=4^3\)
\(\Rightarrow x=4\)
b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)
\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)
\(\frac{x}{10}=\frac{5}{2}\)
\(\Rightarrow x=\frac{5.10}{2}=25\)
2.
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
\(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{3}x=1\\\frac{3}{4}x=\frac{-1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}}\)
vậy \(x=\frac{3}{2}\)hoặc\(x=-\frac{2}{3}\)
\(\left(3+\frac{2}{5}+\frac{3}{4}\right)-\left(2+\frac{1}{5}+\frac{1}{4}\right)=3+\frac{2}{5}+\frac{3}{4}-2-\frac{1}{5}-\frac{1}{4}\)
\(=\left(3-2\right)+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{4}-\frac{1}{4}\right)=1+\frac{1}{5}-\frac{1}{2}=\frac{6}{5}-\frac{1}{2}=\frac{7}{10}\)
Bài làm :
\(\left(3+\frac{2}{5}+\frac{3}{4}\right)-\left(2+\frac{1}{5}+\frac{1}{4}\right)\)
\(=3+\frac{2}{5}+\frac{3}{4}-2-\frac{1}{5}-\frac{1}{4}\)
\(=\left(3-2\right)+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{4}-\frac{1}{4}\right)\)
\(=1+\frac{1}{5}+\frac{1}{2}\)
\(=\frac{17}{10}\)
Học tốt nhé
\(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
\(=\frac{2}{3}-4.\left(\frac{2}{4}+\frac{3}{4}\right)\)
\(=\frac{2}{3}-4.\frac{5}{4}\)
\(=\frac{2}{3}-5\)
\(=-4\frac{1}{3}\)
\(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4\left(\frac{2}{4}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}\)\(\frac{5}{4}\)\(=\frac{2}{3}-\frac{5}{1}=\frac{2}{3}-\frac{15}{3}=\frac{-13}{3}\)