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

\(\left(-3,2\right).\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
=\(-\dfrac{16}{5}.-\dfrac{15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{11}{3}\)

=\(\dfrac{3}{4}+\left(-\dfrac{22}{15}\right):\dfrac{11}{3}\)

=\(\dfrac{3}{4}+\left(-\dfrac{2}{5}\right)\)

=\(\dfrac{7}{20}\)

Đề bài là:Tính các giá trị biểu thức sau ạ

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: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)

\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)

\(\left(-3,2\right).\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

\(=\dfrac{-32}{10}.\dfrac{-15}{64}+\left(\dfrac{8}{10}-\dfrac{34}{15}\right):\dfrac{11}{3}\)

\(=\dfrac{3}{4}+\dfrac{-22}{15}:\dfrac{11}{3}=\dfrac{3}{4}+\dfrac{-2}{5}=\dfrac{7}{20}\)

Cái này bn lầy máy tính ra tính tí là xong thôi

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) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)

= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)

= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)

= \(-\dfrac{5501}{225}\)

\(\left(-3,2\right).\dfrac{15}{64}-\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\\ =-\dfrac{16}{5}.\dfrac{15}{64}-\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{11}{3}\\ =-\dfrac{3}{4}-\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\\ =-\dfrac{3}{4}-\left(-\dfrac{2}{5}\right)=-\dfrac{3}{4}+\dfrac{2}{5}=-\dfrac{7}{20}\)

\(\left(-3,2\right).\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

\(=\left(\dfrac{-16}{5}\right).\dfrac{-15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{11}{3}\)

\(=\left(\dfrac{-16}{5}\right).\dfrac{-15}{64}+\left(\dfrac{-22}{15}\right).\dfrac{3}{11}\)

\(=\dfrac{3}{4}+\left(\dfrac{-2}{5}\right)\)

\(=\dfrac{7}{20}\)

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

\(=4-1-\dfrac{2}{5}\)

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

bấm trên máy tính í là xong ngay í mà

\(=-\dfrac{24}{5}\cdot\dfrac{-10}{64}+\left(\dfrac{4}{5}-\dfrac{19}{15}\right):\dfrac{10}{3}\)

\(=\dfrac{3}{4}-\dfrac{7}{15}\cdot\dfrac{3}{10}=\dfrac{3}{4}-\dfrac{7}{50}=\dfrac{61}{100}\)

\(\left(-4,8\right).\dfrac{-10}{64}+\left(\dfrac{4}{5}-1\dfrac{4}{15}\right):3\dfrac{1}{3}\)

\(=\left(\dfrac{-24}{5}\right).\dfrac{-10}{64}+\left(\dfrac{4}{5}-\dfrac{19}{15}\right).\dfrac{3}{10}\)

\(=\left(\dfrac{-24}{5}\right).\dfrac{-10}{64}+\left(\dfrac{-7}{15}\right).\dfrac{3}{10}\)

\(=\dfrac{3}{4}+\left(\dfrac{-7}{50}\right)\)

\(=\dfrac{61}{100}\)