(0,25 - \(\dfrac{1}{5}\))2
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LG
2
H
15 tháng 5 2023
`4 xx 0,25 xx 5 xx 1/5 xx 2 xx 1/2`
`=(4xx0,25) xx(5xx1/5) xx (2xx1/2)`
`= 1 xx 5/5 xx 2/2`
`= 1 xx 1 xx 1`
`=1`
`@ M``i``n`
12 tháng 9 2021
\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)
12 tháng 9 2021
1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)
LK
28 tháng 4 2018
= \(\left(\dfrac{25}{100}+\dfrac{3}{5}-\dfrac{75}{100}+\dfrac{7}{5}\right):\dfrac{-5}{2}\)
= \(\left(\dfrac{25}{100}+\dfrac{3}{5}_{ }+\dfrac{-75}{100}+\dfrac{7}{5}\right).\dfrac{-2}{5}\)
= \(\left(\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{-3}{4}+\dfrac{7}{5}\right).\dfrac{-2}{5}\)
= \(\left[\left(\dfrac{1}{4}+\dfrac{-3}{4}\right)+\left(\dfrac{3}{5}+\dfrac{7}{5}\right)\right].\dfrac{-2}{5}\)
= \(\left(\dfrac{-2}{4}+\dfrac{10}{5}\right).\dfrac{-2}{5}\)
= \(\left(\dfrac{-10}{20}+\dfrac{40}{20}\right).\dfrac{-2}{5}\)
= \(\dfrac{30}{20}.\dfrac{2}{5}\)
= \(\dfrac{30.2}{20.5}\)
= \(\dfrac{6.1}{10.1}\)
= \(\dfrac{6}{10}=\dfrac{3}{5}\)
LK
28 tháng 4 2018
= (25100+35−75100+75):−52(25100+35−75100+75):−52
= (25100+35+−75100+75).−25(25100+35+−75100+75).−25
= (14+35+−34+75).−25(14+35+−34+75).−25
= [(14+−34)+(35+75)].−25[(14+−34)+(35+75)].−25
= (−24+105).−25(−24+105).−25
= (−1020+4020).−25(−1020+4020).−25
= 3020.253020.25
= 30.220.530.220.5
= 6.110.16.110.1
= 610=35
NL
1
AH
Akai Haruma
Giáo viên
5 tháng 2 2021
Lời giải:
PT $\Leftrightarrow 0,4+0,2x-0,5x=0,25-0,5x+0,25$
$\Leftrightarrow 0,4-0,3x=0,5-0,5x$
$\Leftrightarrow 0,2x=0,1\Rightarrow x=0,5$
DH
13 tháng 6 2017
\(\dfrac{\dfrac{2}{3}+0,25-0,6}{\dfrac{2}{3}-0,25+0,6}:\dfrac{\dfrac{2}{5}-\dfrac{1}{6}+\dfrac{3}{7}}{\dfrac{2}{5}+\dfrac{1}{6}-\dfrac{3}{7}}\)
\(=\dfrac{\dfrac{19}{60}}{\dfrac{61}{60}}:\dfrac{\dfrac{139}{210}}{\dfrac{29}{210}}=\dfrac{19}{61}:\dfrac{139}{29}=\dfrac{551}{8479}\)
Mình không chắc lắm!! Chúc bạn học tốt!!!
OY
21 tháng 7 2021
=\(16.\left(\dfrac{5}{8}-0,25\right):\left(3\dfrac{1}{3}-2\dfrac{1}{4}\right)\)
=\(16.\left(\dfrac{5}{8}-\dfrac{1}{4}\right):\left(\dfrac{10}{3}-\dfrac{9}{4}\right)\)
=\(16.\dfrac{3}{8}:\dfrac{13}{12}\)
=\(6:\dfrac{13}{12}\)
=\(\dfrac{72}{13}\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 10 2023
\(\dfrac{6}{5}\sqrt{1\dfrac{9}{16}}-\left(-\dfrac{3}{4}\right)^2:0,25\)
\(=\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\dfrac{9}{16}:0,25\)
\(=\dfrac{6}{5}\cdot\sqrt{\left(\dfrac{5}{4}\right)^2}-\dfrac{9}{16}:\dfrac{1}{4}\)
\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9\cdot4}{16}\)
\(=\dfrac{6}{4}-\dfrac{9}{4}\)
\(=\dfrac{6-9}{4}\)
\(=-\dfrac{3}{4}\)
19 tháng 12 2020
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
`(0,25 - 1/5)^2`
`=(1/4-1/5)^2`
`=(5/20-4/20)^2`
`=(1/20)^2`
`=1/400`
= (\(\dfrac{15}{10}+\dfrac{2}{10}\))2
= (\(\dfrac{17}{10}\))2
= \(\dfrac{289}{100}\)