Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+-\frac{37}{100}+\frac{1}{8}+-\frac{128}{100}+-\frac{250}{100}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{37}{12}\right)-\left(\frac{37}{100}+\frac{128}{100}+\frac{250}{100}\right)\)
\(=\left(\frac{114}{24}+\frac{3}{24}+\frac{74}{24}\right)-\frac{415}{100}\)
\(=\frac{191}{24}-\frac{415}{100}\)
\(=\frac{457}{120}\)
Tham khảo nha !!!
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+\frac{-37}{100}+\frac{1}{8}+\frac{-32}{25}+\frac{-5}{2}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{-5}{2}\right)+\left(\frac{-37}{100}+\frac{-32}{25}\right)+\frac{37}{12}\)
\(=\frac{19}{8}+\frac{-33}{20}+\frac{37}{12}\)
\(=\frac{29}{40}+\frac{37}{12}\)
\(=\frac{457}{120}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\frac{1}{2}.\left(\frac{20}{15}+\frac{6}{15}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{48}{9}\right)\)
\(=\frac{1}{2}.\frac{26}{15}-\frac{3}{4}.\frac{56}{9}\)
\(=\frac{13}{15}-\frac{14}{3}\)
\(=-\frac{19}{5}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\left(\frac{1}{2}.\frac{4}{3}+\frac{1}{2}.\frac{2}{5}\right)-\left(\frac{3}{4}.\frac{8}{9}+\frac{3}{4}.\frac{16}{3}\right)\)
\(=\left(\frac{2}{3}+\frac{1}{5}\right)-\left(\frac{2}{3}+4\right)\)
\(=\frac{2}{3}+\frac{1}{5}-\frac{2}{3}-4\)
\(=\frac{1}{5}-4\)
\(=\frac{1}{5}-\frac{20}{5}=\frac{-19}{5}\)
\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)
a) Ta có: \(-x+\frac{4}{7}=\frac{1}{3}\)
\(\Leftrightarrow-x=-\frac{5}{21}\)
\(\Rightarrow x=\frac{5}{21}\)
b) Ta có: \(x\div\left(-\frac{1}{3}\right)^2=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^3=-\frac{1}{27}\)
c) \(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{5}\right)^2=\frac{9}{25}\)
\(a.-x+\frac{4}{7}=\frac{1}{3}\)
\(-x=\frac{1}{3}-\frac{4}{7} \)
\(-x=\frac{7}{21}-\frac{12}{21}\)
\(-x=\frac{-5}{21}\)
\(x=\frac{5}{21}\)
\(b.x:\left(\frac{-1}{3}\right)^2=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\left(\frac{-1}{3}\right)^2\)
\(x=\frac{-1}{3}.\frac{-1}{3}.\frac{-1}{3}\)
\(x=\frac{-1}{27}\)
\(c.\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(x=\left(\frac{3}{5}\right)^7:\left(\frac{3}{5}\right)^5\)
\(x=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}.\frac{3}{5}\)
\(x=\frac{9}{25}\)