\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)

câu cuối có sai ko bn?

a) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{3}{4}\right)^2.\left(-1\right)^5\)

\(=\dfrac{8}{27}-\dfrac{9}{16}.\left(-1\right)\)

\(=\dfrac{8}{27}-\left(-\dfrac{9}{16}\right)\)

\(=\dfrac{8}{27}+\dfrac{9}{16}\)

\(=\dfrac{128}{432}+\dfrac{243}{432}\)

\(=\dfrac{371}{432}\)

b) \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(\dfrac{-1}{12}\right)^2\)

\(=12:\dfrac{1}{144}\)

\(=12.144\)

\(=1728\)

c) \(\dfrac{7}{22}:\dfrac{3}{11}+\dfrac{7}{22}:\dfrac{4}{11}\)

\(=\dfrac{7}{22}:\left(\dfrac{3}{11}+\dfrac{4}{11}\right)\)

\(=\dfrac{7}{22}:\dfrac{7}{11}\)

\(=\dfrac{7}{22}.\dfrac{11}{7}\)

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

d) \(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)

\(=\dfrac{12}{35}.5-\dfrac{1}{3}\)

\(=\dfrac{12}{7}-\dfrac{1}{3}\)

\(=\dfrac{36}{21}-\dfrac{7}{21}\)

\(=\dfrac{29}{21}\)

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)$\left(\genfrac{}{}{0ex}{}{1}{2}+1,5\right)\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$2\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{1}{5}:2$

$x=\genfrac{}{}{0ex}{}{1}{10}$
b) $\left(-1\genfrac{}{}{0ex}{}{3}{5}+x\right):\genfrac{}{}{0ex}{}{12}{13}=2\genfrac{}{}{0ex}{}{1}{6}$

$-1\genfrac{}{}{0ex}{}{3}{5}+x=\genfrac{}{}{0ex}{}{13}{6}\cdot \genfrac{}{}{0ex}{}{12}{13}$
$x=2+1\genfrac{}{}{0ex}{}{3}{5}$

$x=3\genfrac{}{}{0ex}{}{3}{5}$
c) $\left(x:2\genfrac{}{}{0ex}{}{1}{3}\right)\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x\cdot \genfrac{}{}{0ex}{}{3}{7}\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x=\genfrac{}{}{0ex}{}{-3}{8}:\genfrac{}{}{0ex}{}{3}{49}$
$x=\genfrac{}{}{0ex}{}{-49}{8}=-6\genfrac{}{}{0ex}{}{1}{8}$
d) $\genfrac{}{}{0ex}{}{-4}{7}\cdot x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}:\left(-1\genfrac{}{}{0ex}{}{2}{3}\right)$

$\genfrac{}{}{0ex}{}{-4}{7}x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{-3}{5}$
$-\genfrac{}{}{0ex}{}{4}{7}x=\genfrac{}{}{0ex}{}{-3}{40}-\genfrac{}{}{0ex}{}{7}{5}x=\genfrac{}{}{0ex}{}{-59}{40}:\genfrac{}{}{0ex}{}{-4}{7}=\genfrac{}{}{0ex}{}{413}{160}=2\genfrac{}{}{0ex}{}{93}{160}$
 

Ai giúp mình làm bước tiếp theo với ạ   

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

\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)

\(=\dfrac{-1}{8}\)

\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{71}{35}\)

\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)

\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)

\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)

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

\(=\dfrac{-433}{70}\)

bạn ơi mk thấy đây đâu có j là hợp lí đâu

1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4

=1/4:1/4-2.-1/8+2

= 1-(-1/4)+2

=1+1/4+2=13/4

2) 3-(-6/7)^0+căn 9 :2

= 3-1+3:2

=3-1+3/2=7/2

3) (-2)^3+1/2:1/8-căn 25 + |-64|

= -8+4-5+64= 55

4) (-1/2)^4+|-2/3|-2007^0

= 1/16+2/3-1

= -13/48

5) = 178/495:623/495-17/60:119/120

= 2/7-2/7=0

6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]

= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]

= -1/2+[1+1+1/2]

= -1/2+5/2=2

Mấy cái dấu chấm đó là  nhân nha bn!

a) $P=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{5}{9}+\frac{1}{12}+\frac{1}{35}$ $=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{5}{9}-\frac{7}{45}\right)+\frac{3}{5}+\frac{1}{35}=1+\frac{4}{5}+\frac{3}{5}+\frac{1}{35}=2\frac{1}{35}$

$\mathrm{a)\; P}=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{5}{9}+\frac{1}{12}+\frac{1}{35}$ 

    P $=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{5}{9}-\frac{7}{45}\right)+\frac{3}{5}+\frac{1}{35}$

P =$1+\frac{4}{5}+\frac{3}{5}+\frac{1}{35}=2\frac{1}{35}$.

b)

$Q\; =\left(5-\frac{3}{4}+\frac{1}{5}\right)-\left(6+\frac{7}{4}-\frac{8}{5}\right)-\left(2-\frac{5}{4}+\frac{16}{5}\right)$

$Q=(5-6-2)+\left(-\frac{3}{4}-\frac{7}{4}+\frac{5}{4}\right)+(\frac{1}{5}-\frac{8}{5}-\frac{16}{5}$)

$Q\; =\; -\left(3+\frac{5}{4}+\frac{23}{5}\right)$ $=-(3+1\frac{1}{4}+4\frac{3}{5}$

$Q\; =-8\frac{17}{20}$.