
\(8^5.64^2.16^3\) Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\) \(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\) \(< =>\frac{-15+12-24}{63}\) \(< =>\frac{-3}{7}\) \(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\) \(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\) \(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\) \(< =>\frac{7}{5}-\frac{4}{7}\) \(< =>\frac{29}{35}\) \(bai2:\) \(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\) \(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\) \(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\) \(< =>x=\frac{3}{5}:\frac{-3}{4}\) \(< =>x=\frac{-4}{5}\) \(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\) \(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\) \(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\) \(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\) Bài 1: a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\) \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) \(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\) \(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\) \(=\frac{29}{35}\) Bài 2: a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\) \(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\) \(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\) \(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\) \(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\) \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) \(x.\frac{10}{3}=\frac{13}{12}\) \(x=\frac{13}{12}:\frac{10}{3}\) \(x=\frac{13}{40}\) a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\) \(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\) \(\frac{3}{7}x=\frac{22}{35}\) \(x=\frac{49}{35}=1,4\) b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\) \(x:\frac{1}{2}=1,5-\frac{1}{4}\) \(x:\frac{1}{2}=\frac{5}{4}\) \(x=\frac{5}{4}.\frac{1}{2}\) \(x=\frac{5}{8}\) Vậy .. \(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\) => \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\) => \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\) Vậy \(x\in\left\{\frac{9}{20}\right\}\) \(b,x+\frac{1}{4}=\frac{4}{3}\) => \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\) Vậy \(x\in\left\{\frac{13}{12}\right\}\) \(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\) => \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\) => \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\) Vậy \(x\in\left\{\frac{25}{42}\right\}\) \(d,\left|x+5\right|-6=9\) => \(\left|x+5\right|=9+6=15\) => \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\) Vậy \(x\in\left\{10;-20\right\}\) \(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\) => \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\) Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\) \(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\) => \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\) => \(\left|x\right|=\frac{1}{6}\) => \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\) Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\) \(g,x^2=16\) => \(\left|x\right|=\sqrt{16}=4\) => \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\) vậy \(x\in\left\{4;-4\right\}\) \(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\) => \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\) => \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) Vậy \(x\in\left\{\frac{5}{6}\right\}\) \(i,3^3.x=3^6\) \(x=3^6:3^3=3^3=27\) Vậy \(x\in\left\{27\right\}\) \(J,\frac{1,35}{0,2}=\frac{1,25}{x}\) => \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\) Vậy \(x\in\left\{\frac{5}{27}\right\}\) \(k,1\frac{2}{3}:x=6:0,3\) => \(\frac{5}{3}:x=20\) => \(x=\frac{5}{3}:20=\frac{1}{12}\) Vậy \(x\in\left\{\frac{1}{12}\right\}\) a,b you cứ tính bt nhé c)\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\) \(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\) \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\) \(=\frac{1}{4}-\frac{1}{11}\) \(=\frac{7}{44}\) d) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\) \(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\) \(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\) \(=5\left(1-\frac{1}{31}\right)\) \(=5.\frac{30}{31}\) \(=\frac{150}{31}\)
b, \(81...">