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Lời giải:
Ta có:
\(\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{13}{4}-\frac{13}{5}+\frac{13}{7}+\frac{13}{11}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{3}{13}\)
\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{17}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)
\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)
\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{13}\)
Dấu "=" chính giữa mình ko biết là dấu + hay - nên mình sẽ làm cả hai TH nhé
TH1: \(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}+\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
\(=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}+\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}+\dfrac{3}{11}=\dfrac{6}{11}\)
Tương tự TH2:
\(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}-\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3}{11}-\dfrac{3}{11}=0\)
Sửa đề:
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(A=\dfrac{3}{5}.\dfrac{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}\right)}\)
\(A=\dfrac{3}{5}+\dfrac{1}{\dfrac{5}{2}}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
Chúc bạn hcọ tốt!!!
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{2}{4}+\dfrac{2}{8}-\dfrac{2}{6}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{6}\right)}{5\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{7}\right)}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
\(A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(A=\dfrac{\dfrac{21}{44}+\dfrac{3}{13}}{\dfrac{20}{77}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{6}+\dfrac{1}{4}}{\dfrac{5}{12}+\dfrac{5}{8}}\)
\(A=\dfrac{\dfrac{645}{1001}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\)
\(\Leftrightarrow A=\left(\dfrac{645}{1001}:\dfrac{645}{1001}\right)+\left(\dfrac{5}{12}:\dfrac{25}{24}\right)\)
\(A=1+\dfrac{2}{5}\)
\(A=\dfrac{7}{5}.\)
Vậy \(A=\dfrac{7}{5}.\)
có gì đó sai sai ở chỗ 5/7 thì phải . mik nghĩ đó 5/4 mới đúng chứ .
\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)
Lời giải:
Ta có:
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\) \(=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)
\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)