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\(\frac{0,25\cdot x}{x}=\frac{5}{6}:0,125\)
\(\Leftrightarrow\frac{\frac{25}{100}\cdot x}{x}=\frac{5}{6}:\frac{125}{1000}\)
\(\Leftrightarrow\frac{\frac{1}{4}\cdot x}{x}=\frac{5}{6}:\frac{1}{8}\)
\(\Leftrightarrow\frac{\frac{1}{4}\cdot x}{x}=\frac{40}{6}\)
\(\Leftrightarrow\frac{1}{4}\cdot x=\frac{40}{6}\cdot x\)
=> x không thỏa mãn
a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)
b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)
c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)
d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
Ta có: \(A=\frac{\frac{3}{11}+1-\frac{3}{7}}{3+\frac{9}{11}-\frac{9}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
\(=\frac{3\left(\frac{1}{11}+\frac{1}{3}-\frac{1}{7}\right)}{9\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{7}\right)}-\frac{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}\)
\(=\frac{3}{9}-\frac{2}{7}=\frac{1}{3}-\frac{2}{7}=\frac{7}{21}-\frac{6}{21}=\frac{1}{21}\)
Vậy \(A=\frac{1}{21}\)
A) (1/5)^5 . 5^5 = 1/5^5 . 5^5 = 5^5 / 5^5 = 1.
B)(0,125)^3 . 512 = (1/8)^3 . 512 = 1/8^3 . 512 = 1/512 . 512 = 1.
C) (0,25)^4 . 1024 = (1/4)^4 . 1024 = 1/4^4 . 1024 = 1/256 . 1024 = 4.
\(\left(0,125\right).\left(-16\right).\left(-\dfrac{8}{9}\right).\left(0,25\right)=\dfrac{1}{8}.\left(-16\right).\left(-\dfrac{8}{9}\right).\dfrac{1}{4}=-2.\left(-\dfrac{2}{9}\right)=\dfrac{4}{9}\)
\(0.25:3x=\frac{5}{6}\)\(:0.125\)
\(\Rightarrow\)\(0.25:3x=\frac{20}{3}\)
\(\Rightarrow\)\(3x=0.25:\frac{20}{3}\)
\(\Rightarrow\)\(3x=\frac{3}{80}\)
\(\Rightarrow\)\(x=\frac{1}{80}\)