Lời giải:

1.

$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$

$=\frac{1}{4^{10}}$

2.

$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$

3.

$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$

giúp mình với ạ

1) \(\left(\dfrac{5}{6}\right)^{10}\cdot\left(\dfrac{3}{10}\right)^{10}\)

\(=\left(\dfrac{5}{6}\cdot\dfrac{3}{10}\right)^{10}\)

\(=\left(\dfrac{1}{4}\right)^{10}\)

2) \(\left(\dfrac{4}{9}\right)^{19}:\left(\dfrac{-12}{35}\right)^{19}\)

\(=\left(\dfrac{4}{9}:\dfrac{-12}{35}\right)^{19}\)

\(=\left(\dfrac{4}{9}\cdot\dfrac{35}{-12}\right)^{19}\)

\(=\left(-\dfrac{35}{27}\right)^{19}\)

3) \(\left(\dfrac{-3}{7}\right)^7:\left(\dfrac{-3}{5}\right)^7\)

\(=\left(\dfrac{-3}{7}:\dfrac{-3}{5}\right)^7\)

\(=\left(\dfrac{-3}{7}\cdot\dfrac{5}{-3}\right)^7\)

\(=\left(\dfrac{5}{7}\right)^7\)

2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)

Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi

0 cần trả lời hết cũng đc

\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)

\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)

\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)

\(=\dfrac{100}{\dfrac{-25}{11}}\)

\(=-44\)

thiếu dấu kìa bạn

ở chỗ 10 và 5/6 ấy

tính giúp mình với

   \(\dfrac{(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)

= \(\dfrac{\left(13-2-10+\left(\dfrac{1}{4}-\dfrac{5}{27}-\dfrac{5}{6}\right)\right).\dfrac{46}{25}+\dfrac{67}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)

= \(\dfrac{\left(1-\dfrac{83}{108}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)

 = \(\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)

=  \(\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{4}}\)

= \(\dfrac{100}{-\dfrac{100}{4}}\)

= - 4 

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(a)\dfrac{4}{5}-\left(\dfrac{-2}{7}\right)-\dfrac{7}{10}\)

\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{-49}{70}\)

\(=\dfrac{27}{70}\)

\(b)\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(=\left(\dfrac{36}{6}+\dfrac{-4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}+\dfrac{-9}{6}\right)-\left(\dfrac{18}{6}+\dfrac{-14}{6}+\dfrac{15}{6}\right)\)

\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

\(=\dfrac{35+\left(-31\right)+\left(-19\right)}{6}\)

\(=-\dfrac{15}{6}=-2\dfrac{1}{2}\)

\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}=\dfrac{4}{5}+\dfrac{2}{7}+\dfrac{-7}{10}=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{-49}{70}=\dfrac{-27}{70}\)

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

= \(6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

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

=\(\left(-2\right)+0+\left(\dfrac{-1}{2}\right)\)

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

a)\(\dfrac{4}{5}\)-\(\left(\dfrac{-2}{7}\right)\)-\(\dfrac{7}{10}\)

=\(\dfrac{56}{70}\)-\(\dfrac{-20}{70}\)-\(\dfrac{49}{70}\)

=\(\dfrac{76}{70}\)-\(\dfrac{49}{70}\)

=\(\dfrac{27}{70}\)

viết p/s như nào vậy bạn