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a) 0,4(3) = \(\frac{4,\left(3\right)}{10}=\frac{4+\frac{1}{3}}{10}=\frac{13}{30}\); 0,6(2) = \(\frac{6,\left(2\right)}{10}=\frac{6+\frac{2}{9}}{10}=\frac{56}{90}=\frac{28}{45}\); 0,5(8) = \(\frac{5,\left(8\right)}{10}=\frac{5+\frac{8}{9}}{10}=\frac{53}{90}\)
Vậy A = \(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{53}{90}}:\frac{2700}{53}\) = \(\frac{13}{30}+\frac{14}{9}-\frac{5}{6}.\frac{90}{53}.\frac{53}{2700}=\frac{13}{30}+\frac{14}{9}-\frac{1}{36}=\frac{353}{180}\)
b) 0,(5) = 5/9; 0,(2) = 2/9
B = \(\left(\frac{5}{9}.\frac{2}{9}\right):\left(\frac{10}{3}.\frac{25}{33}\right)-\left(\frac{2}{5}.\frac{4}{3}\right):\frac{4}{3}\)
B = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)
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a) \(2x\left(x-\frac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy \(x=0;x=\frac{1}{7}\)
b) \(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\\ \left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\\ \left(\frac{5}{10}+\frac{6}{10}\right)x=\frac{-33}{25}\\ \frac{11}{10}x=\frac{-33}{25}\\ x=\frac{-33}{25}:\frac{11}{10}\\ x=\frac{-33.10}{25.11}\\ x=\frac{-6}{5}\)
Vậy x = \(\frac{-6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}:x=\frac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4.3}{9.2}=\frac{2}{3}\\x=\frac{-3}{7}:\frac{-1}{2}=\frac{-3.2}{7.\left(-1\right)}=\frac{6}{7}\end{matrix}\right.\)
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)
b) \(\frac{1}{2}x+\frac{3}{5}x=-\frac{33}{25}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)
\(\Rightarrow x\frac{11}{10}=-\frac{33}{25}\)
\(\Rightarrow x=\left(-\frac{33}{25}\right):\frac{11}{10}=-\frac{33}{25}.\frac{10}{11}=-\frac{6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=0+\frac{4}{9}=\frac{4}{9}\\-\frac{3}{7}:x=0-\frac{1}{2}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4}{9}.\frac{3}{2}=\frac{2}{3}\\x=\left(-\frac{3}{7}\right):\frac{-1}{2}=\left(-\frac{3}{7}\right).\left(-2\right)=\frac{6}{7}\end{matrix}\right.\)
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\(M=\left[0,\left(5\right).0,\left(2\right)\right]:\left(3\frac{1}{3}:\frac{33}{25}\right)-\left[0,4.1,\left(3\right)\right]:1,\left(3\right)=\left(\frac{5}{9}\cdot\frac{2}{9}\right):\left(\frac{10}{3}\cdot\frac{25}{33}\right)-\left(\frac{2}{5}\cdot\frac{4}{3}\right):\frac{4}{3}\)
=\(\frac{10}{81}:\frac{250}{99}-\frac{8}{15}\cdot\frac{3}{4}=\frac{10}{99}\cdot\frac{99}{250}-\frac{2}{5}=\frac{1}{25}-\frac{10}{25}=-\frac{9}{25}\)
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\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)
\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)
\(\Rightarrow\left|x-0,4\right|=6,7\)
\(\Rightarrow x-0,4=\pm6,7\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)
\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)
=> \(\left|x-0,4\right|=\pm6,7\)
Xét hai trường hợp :
TH1 : x - 0,4 = 6,7
=> x = 6,7 + 0,4 = 7,1
TH2 : x - 0,4 = -6,7
=> x = -6,7 + 0,4 =-6,3
\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)
=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)
\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)
=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)
=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)
d, Xét hai trường hợp rồi ra kết quả thôi
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a)\(0,45-\left|1,3-x\right|=0\)
\(\Leftrightarrow\left|1,3-x\right|=0,45-0\)
\(\Leftrightarrow\hept{\begin{cases}1,3-x=0,45\\1,3-x=-0,45\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1,3-0,45\\x=1,3+0,45\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0,85\\x=1,75\end{cases}}\)
Vậy x = 0,85 ; x = 1,75
b) \(\left|3x-5\right|-\frac{1}{7}=\frac{1}{3}\)
\(\Leftrightarrow\left|3x-5\right|=\frac{1}{3}+\frac{1}{7}\)
\(\Leftrightarrow\left|3x-5\right|=\frac{10}{21}\)
\(\Leftrightarrow\hept{\begin{cases}3x-5=\frac{10}{21}\\3x-5=-\frac{10}{21}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}3x=\frac{10}{21}+5\\3x=-\frac{10}{21}+5\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}3x=\frac{115}{21}\\3x=\frac{95}{21}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{115}{63}\\x=\frac{95}{63}\end{cases}}\)
Vậy x = .........................
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a) Vì \(0,\left(3\right)=\dfrac{3-0}{9}=\dfrac{3}{9}=\dfrac{1}{3}\) và \(-0,4\left(2\right)=-\dfrac{42-4}{90}=-\dfrac{38}{90}=-\dfrac{19}{45}\) nên:
\(0,\left(3\right)+3\dfrac{1}{3}-0,4\left(2\right)=\dfrac{1}{3}+\dfrac{10}{3}-\dfrac{19}{45}=\dfrac{11}{3}-\dfrac{49}{45}\)
\(=\dfrac{165-19}{45}=\dfrac{146}{45}\)
b) Vì \(0,\left(5\right)=\dfrac{5-0}{9}=\dfrac{5}{9}\) và \(0,\left(2\right)=\dfrac{2-0}{9}=\dfrac{2}{9}\) nên:
\(\left[0,\left(5\right).0,\left(2\right)\right]:\left(3\dfrac{1}{3}:\dfrac{33}{25}\right)=\left(\dfrac{5}{9}.\dfrac{2}{9}\right):\left(\dfrac{10}{3}.\dfrac{25}{33}\right)=\dfrac{10}{81}:\left(\dfrac{110.25}{33}\right)\)
\(=\dfrac{10}{81}.\dfrac{33}{110.25}=\dfrac{3}{81.25}=\dfrac{1}{27.25}=\dfrac{1}{675}\)