Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
B=\(\frac{12+\frac{4}{3}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12+\frac{12}{9}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{9}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}:\frac{4\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
=\(4:\frac{4}{5}\)
=\(5\)
chơ một ti nhá