\(=\frac{5}{6}x\left(\frac{7}{12}+\frac{5}{12}\right)+\frac{1}{6}=\frac{5}{6}x1+\frac{1}{6}=1\)
a) \(\frac{5}{6}x\frac{7}{12}+\frac{5}{12}x\frac{5}{6}+\frac{1}{6}=\frac{5}{6}x\left(\frac{7}{12}+\frac{5}{12}\right)+\frac{1}{6}=\frac{5}{6}x1+\frac{1}{6}=\frac{5}{6}+\frac{1}{6}=1\)
b) \(\left(\frac{8}{15}x\frac{3}{4}x\frac{5}{4}\right):\frac{7}{8}=\frac{4x2x3x5}{5x3x4x4}x\frac{8}{7}=\frac{2}{4}x\frac{8}{7}=\frac{2x8}{4x7}=\frac{2x4x2}{4x7}=\frac{4}{7}\)
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