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\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)
\(A=(1\frac{1}{5}+\frac{17}{24}+\frac{29}{32}+\frac{1}{4}+\frac{7}{24}+\frac{3}{32})\div4\)
\(A=\left\{1\frac{1}{5}+\frac{1}{4}+(\frac{17}{24}+\frac{7}{24})+(\frac{29}{32}+\frac{3}{32})\right\}\div4\)
\(A=\left\{\frac{29}{20}+1+1\right\}\div4\)
\(A=\frac{69}{20}\div4\)
\(A=\frac{69}{80}=0,8625\)
a=(6/5+17/24+29/32+1/4+7/24+3/32):4
a=[6/5+(17/24+7/24)+(29/32+3/32+1/4)]:4
a=[6/5+1+1+1/4):4
a=[24/20+20/20+20/20+5/20]:4
a=[44/20+25/20]:4
a=69/20:4
a=69/80
a=
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)
\(D=\dfrac{1}{3}+\dfrac{2}{3}\)
\(D=1\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)
D = \(\dfrac{3}{3}\)
D = 1