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\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(\Rightarrow1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(\Rightarrow\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=1\)
\(\Rightarrow\frac{43}{8}+x-\frac{173}{24}=\frac{53}{3}\)
\(\Rightarrow\frac{43}{8}+x=\frac{53}{3}+\frac{173}{24}\)
\(\Rightarrow\frac{43}{8}+x=\frac{199}{8}\)
\(\Rightarrow x=\frac{199}{8}-\frac{43}{8}\)
\(\Rightarrow x=\frac{156}{8}=\frac{39}{2}\)
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(< =>1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(< =>1\left(\frac{43}{8}+x-\frac{173}{24}\right)=0\)
\(< =>\frac{43}{8}+x-\frac{173}{24}=0\)
\(< =>\frac{43}{8}+x=\frac{173}{24}\)
\(< =>x=\frac{173}{24}-\frac{43}{8}\)
\(< =>x=\frac{11}{6}\)
\(a,\left(16.23+16.77\right)-\left(5.30-5.20\right)\)
\(=16.\left(23+77\right)-5.\left(30-20\right)\)
\(=16.100-5.10\)
\(=1600-50\)
\(=1550\)
\(b,8\frac{1}{2}\div\frac{17}{5}+\frac{6}{8}\div3\frac{2}{3}\)
\(=\frac{17}{2}.\frac{5}{17}+\frac{3}{4}\div\frac{11}{3}\)
\(=\frac{17}{2}.\frac{5}{17}+\frac{3}{4}.\frac{3}{11}\)
\(=\frac{5}{2}+\frac{9}{44}\)
\(=\frac{110}{44}+\frac{9}{44}\)
\(=\frac{119}{44}\)
\(a,=16.100-5.10=1600-50=1550\)
\(b,8\frac{1}{2}:\frac{17}{5}+\frac{6}{8}:3\frac{2}{3}=\frac{17}{2}.\frac{5}{17}+\frac{6}{8}:\frac{11}{3}=\frac{5}{2}+\frac{18}{88}=\frac{220}{88}+\frac{18}{88}=\frac{238}{88}=2\frac{31}{44}\)