d)2/11⋅−5/4+−9/11⋅5/4+1/34 f)3,4⋅(−23,68)−3,4⋅45,12+(−31,2)⋅3,4
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3,4 x (-23,68) - 3,4 x 45,12 + (-31,2) x 3,4
3,4 . [ (-23,68) - 45,12,+ (-31,2)]
=3,4 . -100
=-340
a: =-12,45+12,45+23,4=23,4
b: =32,18-32,18-14,6+14,6+4,125=4,125
c: =4,5(-12,25-17,75)
=-4,5*30=-135
d: =3,4(-23,68-45,12-31,2)
=3,4*(-100)=-340
\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)
\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)
\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)
\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)
\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)
\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)
\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)
\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)
\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)
\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)
\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)
\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)
\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)
\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)
\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)
\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)
\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)
\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)
\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)
\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)
1: =290-40=250
2: =72,69+18,47-8,47-22,69=50+10=60
3: =2,07-7,64+8,79+1,21-7,36
=10-15+2,07
=-5+2,07
=-2,93
6: =3,58(24,25+75,55)
=100*3,58
=358
1.
a)=1/3-[(-5/4)-5/8]
=1/3-(-15/8)=53/24
b)=5/9:(-3/22)+5/9:(-3/5)
=5/9*22/-3+5/9*5/-3=-110/27+-25/27=5
2
a)Ta có 339<340=920<1120<1121
nên 339<1121
b)Ta có /3,4-x/ lớn hơn hoặc bằng 0 Với mọi x thuộc R
=> -/3,4-x/ bé hơn hoặc bằng 0 Với mọi x thuộc R
=> 0,5-/3,4-x/ bé hơn hoặc bằng 0,5 Với mọi x thuộc R
Dấu = xảy ra khi 3,4-x=0
=>x=3,4
Vậy GTLN của A = 0,5 khi x=3,4
a) Ta có:
\(\dfrac{7}{11}=0,\left(63\right)\)
\(\dfrac{8}{-13}=-0,\left(615384\right)\)
\(\dfrac{37}{19}\approx1,95\)
\(\dfrac{-5}{13}=-0,\left(384615\right)\)
\(\Rightarrow-0,\left(615384\right)< -0,\left(384615\right)< 0< 0,\left(63\right)< 1,95\)
\(\Rightarrow\dfrac{8}{-13}< \dfrac{-5}{13}< 0< \dfrac{7}{11}< \dfrac{37}{19}\)
b) Ta có:
\(\dfrac{-9}{17}\approx-0,53\)
\(-\dfrac{6}{5}=-1,2\)
\(\dfrac{7}{-17}\approx-0,41\)
\(\Rightarrow-3,4< -1,2< -1,02< -0,53< -0,41< 0\)
\(\Rightarrow-3,4< -\dfrac{6}{5}< -1,02< \dfrac{-9}{17}< \dfrac{7}{-17}< 0\)
a: 0<7/11
7/11<1
1=19/19<37/19
=>0<7/11<37/19
mà -8/13<-5/13<0
nên -8/13<-5/13<0<7/11<37/19
b: -1<-9/17<-7/17<0
-3,4<-1,2=-6/5<-1,02<-1
Do đó: -3,4<-6/5<-1,02<-1<-9/17<-7/17<0
a: =22+25,5=47,5
b: =-9,207-1,505+3,8-2,8=1-10,712=-9,712
c: \(=2,07+3,005-12,005+4,23=6,3-9=-2,7\)
e: =3,58(24,45+75,55)=3,58*100=358