Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{2}{4}+\dfrac{2}{8}-\dfrac{2}{6}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{6}\right)}{5\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{7}\right)}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
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!!!
Lời giải:
Ta có:
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\) \(=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
\(\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}\)+\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{3}\right)}\)
=\(\frac{3}{5}\)+\(\frac{1}{\frac{5}{2}}\)
=\(\frac{3}{5}\)+\(\frac{2}{5}\)
=1 !!!
\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}+\dfrac{\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}}\)
\(=\dfrac{2\cdot\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\cdot\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}=1\)
\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}+\dfrac{\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}}\\ =\dfrac{2\cdot\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\cdot\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}\right)}{3\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\\ =\dfrac{2}{3}+\dfrac{1}{3}\\ =\dfrac{3}{3}\\ =1\)