Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)
b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
\(\left(-3,2\right).\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
\(=\dfrac{-32}{10}.\dfrac{-15}{64}+\left(\dfrac{8}{10}-\dfrac{34}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}:\dfrac{11}{3}=\dfrac{3}{4}+\dfrac{-2}{5}=\dfrac{7}{20}\)
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)
=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)
=\(\left[80+9\right]=89\)
\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)
\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)
\(=80+\left(3^5\div3^2\right)\)
\(=80+3^3\)
\(=80+27\)
\(=107\)
e) \(\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}.\dfrac{5}{7}+2\dfrac{3}{5}\)
= \(\dfrac{-3}{5}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}.1+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}+\dfrac{13}{5}\)
= 2
\(\left(-3,2\right).\dfrac{15}{64}-\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\\ =-\dfrac{16}{5}.\dfrac{15}{64}-\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{11}{3}\\ =-\dfrac{3}{4}-\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\\ =-\dfrac{3}{4}-\left(-\dfrac{2}{5}\right)=-\dfrac{3}{4}+\dfrac{2}{5}=-\dfrac{7}{20}\)
\(\left(-3,2\right).\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
\(=\left(\dfrac{-16}{5}\right).\dfrac{-15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{11}{3}\)
\(=\left(\dfrac{-16}{5}\right).\dfrac{-15}{64}+\left(\dfrac{-22}{15}\right).\dfrac{3}{11}\)
\(=\dfrac{3}{4}+\left(\dfrac{-2}{5}\right)\)
\(=\dfrac{7}{20}\)
\(=-\dfrac{24}{5}\cdot\dfrac{-10}{64}+\left(\dfrac{4}{5}-\dfrac{19}{15}\right):\dfrac{10}{3}\)
\(=\dfrac{3}{4}-\dfrac{7}{15}\cdot\dfrac{3}{10}=\dfrac{3}{4}-\dfrac{7}{50}=\dfrac{61}{100}\)
Bạn học lớp nào trường gì mà bt về nhà môn Toán giống của bọn mik zậy ?
Mik học lớp 6C trường THCS Đại An