\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
= \(1+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
= \(1+\frac{3}{4}\)
= \(\frac{4}{4}+\frac{3}{4}\)
= \(\frac{7}{4}\)
HỌC TỐT
Ta có: \(B=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=1-\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=1-\frac{3}{4}=\frac{1}{4}\)
cho em hỏi tại sao trên cộng dưới trừ ạ