lớp 1 mà cậu

4.24.5²-(3³.18+3³.12)

=4.24.25-[27.(18+12)]

=(4.25).24-[27.30]

=100.24-810

=2400-810

=1590

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)

\(=-\dfrac{37}{16}\)

\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)

\(=\dfrac{5}{17}+\dfrac{-3}{17}\)

\(=\dfrac{2}{17}\)

\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)

\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)

\(=2-\dfrac{7}{30}\)

\(=\dfrac{53}{30}\)

\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)

\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)

\(=\dfrac{15}{34}\)

Ý a anh chép sai đề bài nên làm sai rùi kìa!~

có đúng ko

dài thế này làm đến tết chắc vẫn chưa xong

Bạn đăng từ bài thôi , nhiều quá sao làm nổi

Bài 1: Tính

\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)

\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)

\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)

\(=\dfrac{5}{8}.\dfrac{-4}{15}\)

\(=\dfrac{-1}{6}\)

\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)

\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)

\(=\dfrac{-21}{40}-\dfrac{3}{4}\)

\(=\dfrac{-51}{40}\)

\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)

\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)

\(=\dfrac{4}{6}\)

\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)

\(=\dfrac{4}{3}-1\)

\(=\dfrac{1}{3}\)

\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)

\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)

\(=1:\dfrac{1}{5}\)

\(=5\)

\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)

\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)

\(=3+\dfrac{11}{7}\)

\(=3\dfrac{11}{7}=\dfrac{32}{7}\)

\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)

\(=1:\dfrac{19}{5}\)

\(=\dfrac{5}{19}\)

\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)

\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)

\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)

\(=1:\left(\dfrac{2}{3}+1\right)\)

\(=1:\dfrac{5}{3}\)

\(=\dfrac{3}{5}\)
\(\text{9)}\)

\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)

\(=\dfrac{330875}{1507764}\)

\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)

\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)

\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)

Gigachad cảm ơn bạn sai rồi

a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)

\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)

\(A=\dfrac{-1}{10.2}\)

\(A=\dfrac{-1}{20}\)

b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)

\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)

\(B=\dfrac{2}{3.3}\)

\(B=\dfrac{2}{9}\)

c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)

\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)

\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)

\(C=\dfrac{2}{3}+\dfrac{2}{3}\)

\(C=\dfrac{4}{3}\)

d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)

\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)

\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)

\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)

\(D=\dfrac{-57}{57}=-1\)

e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)

\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)

\(E=\dfrac{27}{182}-\dfrac{27}{182}\)

\(E=0\)