x - 5/6 : 1 trên1/3 =0,125
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\(x-\frac{5}{6}:1\frac{1}{3}=0,125\)
\(x-\frac{5}{6}:\frac{4}{3}=0,125\)
\(x-\frac{5}{6}\cdot\frac{4}{3}=\frac{1}{8}\)
\(x-\frac{10}{9}=\frac{1}{8}\)
\(x=\frac{10}{9}+\frac{1}{8}\)
\(x=\frac{89}{9}\)
Phương Boice nhìn bài soyeon_Tiểu bàng giải nên lm sai => cả 2 người đều lm sai, ai thấy đúng thì k
\(x-\frac{5}{6}:1\frac{1}{3}=0,125\)
\(x-\frac{5}{6}:\frac{4}{3}=\frac{1}{8}\)
\(x-\frac{5}{6}.\frac{3}{4}=\frac{1}{8}\)
\(x-\frac{5}{8}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{8}\)
\(x=\frac{6}{8}=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
a) \(\left(\dfrac{1}{4}x\right):3=\dfrac{5}{6}:\dfrac{1}{8}\\ \left(\dfrac{1}{4}x\right):3=\dfrac{20}{3}\\ \dfrac{1}{4}x=\dfrac{20.3}{3}\\ \dfrac{1}{4}x=20\\ x=80\)
a: Ta có: \(\dfrac{0.25x}{3}=\dfrac{5.6}{0.125}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=134.4\)
hay x=537,6
b: Ta có: \(\dfrac{2.5}{7.5}=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{5}\)
\(A=\frac{\frac{1}{99}+\frac{1}{98}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{\left(\frac{1}{99}+1\right)+\left(\frac{1}{98}+1\right)+...+\left(\frac{98}{2}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=\frac{100\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(A=100\)
\(a,x-\frac{5}{6}:1\frac{1}{6}=0,125\)
\(x-\frac{5}{6}:\frac{7}{6}=\frac{1}{8}\)
\(x-\frac{5}{7}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{7}\) \(x=\frac{47}{56}\)
\(b,\left(1-\frac{2}{10}+x+\frac{1}{5}\right):\left(1\frac{1}{3}-\frac{2}{3}+3\frac{1}{3}\right)-1=1\frac{1}{2}\)
\(\left(1-\frac{1}{5}+x+\frac{1}{5}\right):\left(\frac{4}{3}-\frac{2}{3}+\frac{10}{3}\right)-1=\frac{3}{2}\)
\(\left(\frac{4}{5}+x+\frac{1}{5}\right):4=\frac{3}{2}+1\)
\(\left(1+x\right):4=\frac{5}{2}\)
\(1+x=\frac{5}{2}.4\)
\(1+x=10\)
\(x=10-1\)
\(x=9\)
\(x-\frac{5}{6}\div\frac{1}{\frac{1}{3}}=0,125\)
\(x-\frac{5}{6}\div3=\frac{1}{8}\)
\(x-\frac{5}{18}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{18}\)
\(x=\frac{29}{72}\)
\(x-\frac{5}{6}:\frac{1}{\frac{1}{3}}=0,125\)
\(x-\frac{5}{6}:\frac{1}{\frac{1}{3}}=\frac{1}{8}\)
\(x-\frac{5}{6}:3=\frac{1}{8}\)
\(x-\frac{5}{18}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{18}\)
\(x=\frac{29}{72}\)
Vậy x = \(\frac{29}{72}\)