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a)\(\frac{-1}{3}.\frac{141}{17}-13.\frac{-1}{17}\)
\(=\frac{141}{3}.\frac{-1}{17}-13.\frac{-1}{17}\)
\(=47.\frac{-1}{17}-13.\frac{-1}{3}\)
\(=\frac{-1}{17}.\left(47-13\right)\)
\(=\frac{-1}{17}.33=....\)(tự tính)
\(a,\)\(\frac{-1}{3}.\frac{141}{17}-13.\frac{-1}{7}\)
\(=\frac{-1.141}{3.17}+13.\frac{1}{17}\)
\(=\frac{-141}{17}.\frac{1}{7}+13.\frac{1}{7}\)
\(=\frac{1}{7}\left(-47+13\right)\)
\(=\frac{-34}{7}\)
\(b,\)\(\frac{-9}{16}.\frac{13}{3}-\left(-\frac{3}{4}\right)^2.\frac{19}{3}\)
\(=\left(\frac{3}{4}\right)^2.\frac{-13}{3}-\left(-\frac{9}{16}\right).\frac{19}{3}\)
\(=\frac{9}{16}.\frac{-13}{3}+\frac{9}{16}.\frac{19}{3}\)
\(=\frac{9}{16}\left(\frac{-13}{3}+\frac{19}{3}\right)=\frac{9}{16}.2=\frac{9}{8}\)
a3.52-16:22 b 23.17-23.14 c15-141+59.15=sai đề e 10-[30-(5-1)2]=sai đề
=3.25-16:4 =23.(17-14) d 17.85+15.17-120
=75-4 =23.3 =17.(85+15)-120
=71 =8.3=24 =17.100-120
=1700-120=1580
1. a, M = -\(\dfrac{1}{3}.\dfrac{141}{17}-\dfrac{39}{3}.\left(-\dfrac{1}{17}\right)\)
= -\(\dfrac{1}{17}.\dfrac{141}{3}-\dfrac{39}{3}.\left(-\dfrac{1}{17}\right)\)
= -\(\dfrac{1}{17}\left(\dfrac{141}{3}-\dfrac{39}{3}\right)\)
= -\(\dfrac{1}{17}.34\)
= -2
@Lê Thị Hồng Ngát
1. b, \(\dfrac{3}{4}+\dfrac{1}{4}x=7\)
<=> \(\dfrac{1}{4}x=\dfrac{25}{4}\)
<=> x = 25
@Lê Thị Hồng Ngát
160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580
15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216
160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580 15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216