Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = \(\dfrac{-3}{14}\) : \(\dfrac{5}{7}\)
\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = - \(\dfrac{3}{10}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{3}{10}\) + \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{19}{10}\)
\(x\) = \(\dfrac{19}{10}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{19}{6}\)
\(\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}:\dfrac{5}{7}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{10}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{10}+\dfrac{11}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{19}{10}\)
\(\Rightarrow x=\dfrac{19}{10}:\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{19}{6}\)
a) https://hoc247.net/hoi-dap/toan-7/thuc-hien-phep-tinh-5-5-27-7-23-0-5-5-27-16-23-faq218258.html
b) https://hoidap247.com/cau-hoi/1507941
B=(\(\dfrac{1}{5}\)-\(\dfrac{1}{5}\))+(\(\dfrac{3}{7}\)-\(\dfrac{3}{7}\))+(\(\dfrac{5}{9}\)-\(\dfrac{5}{9}\))+(\(\dfrac{2}{11}\)-\(\dfrac{2}{11}\))+(\(\dfrac{7}{13}\)-\(\dfrac{7}{13}\))+\(\dfrac{9}{16}\)
B=\(\dfrac{9}{16}\)
A=(\(\dfrac{1}{3}-\dfrac{1}{3}\))\(+\left(\dfrac{3}{5}+\left(\dfrac{-3}{5}\right)\right)+\left(\dfrac{-5}{7}+\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)\)\(+\left(\dfrac{-11}{13}-\dfrac{9}{11}\right)\)
A\(=0+0+0+0+\dfrac{-238}{143}\)
A\(=\dfrac{-238}{143}\)
\(B=\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{4}\right)+\left(1+\dfrac{1}{8}\right)+\left(1+\dfrac{1}{32}\right)+\left(1+\dfrac{1}{64}\right)-7\)
\(B=\left(1+1+1+1+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)-7\)
\(B=6+\dfrac{63}{64}-7\)
\(B=-1+\dfrac{63}{64}\)
\(B=\dfrac{-1}{64}\)
bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)
\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)
\(=\dfrac{-27.1.1}{5.1.9}\)
\(=\dfrac{-27}{45}\)
\(=\dfrac{-9}{15}\)
b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{14}+\dfrac{5}{7}+1\\ =\dfrac{5}{7}\cdot\left(\dfrac{-2}{11}+\dfrac{-9}{14}+1\right)+1\\ =\dfrac{5}{7}\cdot\dfrac{27}{154}+1\\ =\dfrac{135}{1078}+1\\ =\dfrac{1213}{1078}\)