125%.\(\left(\dfrac{-1}{2}\right)^2\):\(\left(1\dfrac{5}{16}-1,5\right)\)+\(2018^0\)
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\(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{24}{16}\right)+1=\dfrac{5}{16}\cdot\dfrac{16}{-3}+1=\dfrac{-5}{3}+1=-\dfrac{2}{3}\)
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 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
`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...
\(125\%\cdot\left(-\dfrac{1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2018^0\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)+1\)
\(=\dfrac{5}{16}:\left(-\dfrac{3}{16}\right)+1\)
\(=\dfrac{5}{16}\cdot\left(-\dfrac{16}{3}\right)+1\)
\(=-\dfrac{5}{3}+1\)
\(=-\dfrac{5}{3}+\dfrac{3}{3}\)
\(=-\dfrac{2}{3}\)
`125%*( (-1)/2 )^2 :(1 5/16 -1,5 )+2018^0`
`=1,25 * 1/4 :(21/16-3/2)+1`
`=1,25*1/4:(21/16-24/16)+1`
`=1,25*1/4:(-3)/16+1`
`=5/16*16/(-3)+1`
`=-5/3+1`
`=-5/3+3/3`
`=-2/3`