a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

\(a,-\dfrac{3}{5}.y=\dfrac{21}{10}\)

\(y=\dfrac{21}{10}:\dfrac{-3}{5}=\dfrac{-7}{2}\)

\(b,y:\dfrac{3}{8}=-1\dfrac{31}{33}\)

\(y=-1\dfrac{31}{33}.\dfrac{3}{8}=\dfrac{-8}{11}\)

Vậy \(y=-\dfrac{8}{11}\)

\(c,1\dfrac{2}{5}.y+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\Rightarrow1\dfrac{2}{5}y=-\dfrac{4}{5}-\dfrac{3}{7}=\dfrac{-43}{35}\)

\(\Rightarrow y=\dfrac{-43}{35}:1\dfrac{2}{5}=\dfrac{-43}{49}\)

\(d,-\dfrac{11}{12}.y+0,25=\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{11}{12}.y=\dfrac{5}{6}-0,25=\dfrac{7}{12}\)

\(\Rightarrow y=\dfrac{7}{12}:\dfrac{-11}{12}=\dfrac{-7}{11}\)

a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)

\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)

b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)

\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)

a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)

b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)

\(A\cdot\left(1-\dfrac{1}{4}\right)\cdot\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)=1\dfrac{3}{5}\)

=>\(A\cdot\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)=\dfrac{8}{5}\)

=>\(A\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}=\dfrac{8}{5}\)

=>\(A\cdot\dfrac{1}{5}\cdot\dfrac{6}{2}=\dfrac{8}{5}\)

=>\(A\cdot3=8\)

=>A=8/3

a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)

b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)

c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)

a: =

b: =

c: =

\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)

\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)

\(=0\)

\(\dfrac{5}{3}\times\dfrac{15}{8}-\dfrac{9}{8}\times\dfrac{5}{8}-\dfrac{5}{3}\)

\(=\dfrac{5}{3}\times\left(\dfrac{15}{8}-\dfrac{9}{8}-1\right)\)

\(=\dfrac{5}{3}\times\left(\dfrac{15}{8}-\dfrac{9}{8}-\dfrac{8}{8}\right)\)

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

\(=-\dfrac{5}{12}\)

a)

\(\dfrac{3}{5}\times\dfrac{7}{11}\times\dfrac{5}{3}\times11\\ =\left(\dfrac{3}{5}\times\dfrac{5}{3}\right)\times\left(\dfrac{7}{11}\times11\right)\\ =\dfrac{3\times5}{5\times3}\times\dfrac{7\times11}{11}\\ =1\times7\\ =7\)

b)

\(\dfrac{3}{8}\times\dfrac{2}{7}+\dfrac{5}{7}\times\dfrac{3}{8}\\ =\dfrac{3}{8}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\\ =\dfrac{3}{8}\times\dfrac{7}{7}\\ =\dfrac{3}{8}\times1=\dfrac{3}{8}\)

a) $\frac{2}{5} \times \frac{3}{8} \times \frac{3}{4} = \frac{{2 \times 3 \times 3}}{{5 \times 8 \times 4}} = \frac{{18}}{{160}} = \frac{9}{{80}}$

b) $\frac{1}{3} \times \frac{1}{6} \times \frac{1}{9} = \frac{{1 \times 1 \times 1}}{{3 \times 6 \times 9}} = \frac{1}{{162}}$

c) $\frac{3}{4}:\frac{1}{5}:\frac{7}{8} = \frac{3}{4} \times \frac{5}{1} \times \frac{8}{7} = \frac{{3 \times 5 \times 8}}{{4 \times 1 \times 7}} = \frac{{120}}{{28}} = \frac{{30}}{7}$

d) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$