`125% . ((-1)/2)^2 : ( 1 5/16 - 1,5)+2018^0`

`= 125/100 . 1/4 : ( 21/5 - 3/2 ) + 1`

`= 5/4 . 1/4 : 27/10 + 1`

`= 5/4 . 1/4 . 10/27 +1`

`= 5/16 . 10/27 +1`

`=25/215 +1`

`=241/216`

125%⋅(−21​)2:(1165​−1,5)+20180
=54⋅14:(2116−32)+1=45​⋅41​:(1621​−23​)+1
=516:(−316)+1=165​:(−163​)+1
=516⋅(−163)+1=165​⋅(−316​)+1
=−53+1=−35​+1
=−53+33=−35​+33​
=−23=−32​
 

\(125\%.\left(-\dfrac{1}{2}\right)^2:\left(1\dfrac{5}{16}-1.5\right)+2018^0\)

\(=\dfrac{5}{4}.\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)+1\)

\(=\dfrac{5}{16}:-\dfrac{3}{16}+1\)

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

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

\(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{24}{16}\right)+1=\dfrac{5}{16}\cdot\dfrac{16}{-3}+1=\dfrac{-5}{3}+1=-\dfrac{2}{3}\)

Bài 1: 

a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)

c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

`a)4/5+5 1/2 xx (4,5-2)+7/10`

`=4/5+11/2*2,5+7/10`

`=0,8+2,2+0,7`

`=3+0,7=3,7`

`b)125%xx 17/4:(1 5/16-0,5)+2008`

`=1,25xx4,25:13/16+2008`

`=85/13+2008`

`=2014 7/13`

`c)5/11+(16/11+1)`

`=5/11+1+5/11+1`

`=2+10/11=32/11`

`d)3/17+11/4+5/8+14/17+3/8`

`=3/17+14/17+5/8+3/8+11/4`

`=1+1+11/4`

`=19/4`

a) 

\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)

<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)

<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)

b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)

= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)

c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)

= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)

d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)

= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

dài lắm mk ngại viết bạn cứ tính lần lượt là xong ngay

\(\dfrac{2}{3}\times\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\times\dfrac{1}{3}\) 

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

= \(\dfrac{1}{3}\times\dfrac{19}{15}-\dfrac{2}{3}\) 

= \(\dfrac{19}{45}-\dfrac{2}{3}\) 

=\(-\dfrac{11}{45}\)

\(\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}\) 

(\(\dfrac{10}{3}+\dfrac{5}{2}):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}\)

\(\dfrac{35}{6}:\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\) 

\(\dfrac{36}{5}:-\dfrac{31}{30}-\dfrac{11}{31}\) 

\(-\dfrac{216}{31}\) - \(\dfrac{11}{31}\) 

-227/31

ai giúp mìn vứi ❤

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

a/ \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{-5}{12}\)

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

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{20}\)

\(\Leftrightarrow x=\dfrac{29}{10}\)

Vậy ...

b/ \(\left(4x-3\right)\left(\dfrac{5}{4}x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{8}{5}\end{matrix}\right.\)

Vậy .....

c/ \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)

\(\Leftrightarrow\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=-\dfrac{9}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=-\dfrac{19}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{38}{21}\end{matrix}\right.\)

Vậy ......

d/ \(\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)

\(\Leftrightarrow\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)

\(\Leftrightarrow\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{9}{10}\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy ...

a. \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)

\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{6}-\dfrac{-5}{12}\)

\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{4}\)

\(\dfrac{3}{6}x=\dfrac{5}{4}+\dfrac{1}{5}\)

\(\dfrac{3}{6}x=\dfrac{29}{20}\)

\(x=\dfrac{29}{20}:\dfrac{3}{6}\)

\(x=\dfrac{29}{10}\)

Vậy...

b. \(\left(4x-3\right).\left(\dfrac{5}{4}x+2\right)=0\)

\(\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-8}{5}\end{matrix}\right.\)

Vậy ...

c. \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)

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

\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{-9}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=\dfrac{-19}{12}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{-38}{21}\end{matrix}\right.\)

Vậy...