Bài 1:Tính
a,\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}\)\(+\frac{3}{7}\)
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\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)
\(=\frac{4\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)
\(=\frac{4}{7}+\frac{3}{7}\)
\(=\frac{7}{7}=1\)
\(a)\)\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)
\(=\)\(\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)
\(=\)\(\frac{4}{7}+\frac{3}{7}\)
\(=\)\(1\)
Mk chỉ biết câu a thôi
a) \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}\)
= \(\frac{5\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}{15\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}\)
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
Chúc bạn học tốt
a) \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}=\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{\left(10+5\right)}{7}+\frac{\left(10+5\right)}{9}-\frac{\left(10+5\right)}{11}}\)
\(=\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\left(\frac{10}{7}+\frac{5}{7}\right)+\left(\frac{10}{9}+\frac{5}{9}\right)-\left(\frac{10}{11}+\frac{5}{11}\right)}=\frac{0}{\frac{10}{7}+\frac{10}{9}-\frac{10}{11}}=0\)
b) \(\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{1}{73}+\frac{4}{73}-\frac{5}{115}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{1}{73}+\frac{4}{73}-\frac{4}{115}+\frac{1}{115}}\)
\(=\frac{4}{5+\frac{1}{73}-\frac{1}{115}}=0\)
tth làm dài quá, làm gọn hơn nè
a, \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}=\frac{5\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}{15\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}=\frac{5}{15}=\frac{1}{3}\)
b, \(\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{5}{115}}=\frac{4\left(1+\frac{1}{73}-\frac{1}{115}\right)}{5\left(1+\frac{1}{73}-\frac{1}{115}\right)}=\frac{4}{5}\)
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{\left(-7\right).\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{-1}{7}\)
\(A=\frac{21}{35}+\frac{-5}{35}\)
\(A=\frac{16}{35}\)
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-\frac{7}{2}+\frac{7}{3}-\frac{7}{4}+\frac{7}{5}}\)
\(=\frac{3}{5}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
=3/5+(1/-7)
=3/5-1/7
=16/35
\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)
\(=\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)
\(=\frac{4}{7}+\frac{3}{7}=1\)
\(a)\)\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)
\(=\)\(\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)
\(=\)\(\frac{4}{7}+\frac{3}{7}\)
\(=\)\(1\)