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

Ta có : x⁴ - 12x³ + 12x² - 12x + 111 

= x³(x - 12) + 12x(x - 1) + 111

Thay x = 11 vào => 11³(11 - 12) + 12.11.(11 - 1) + 111

= 11³ + 120.11 + 111

= 121.11 + 120.11 + 111

= 11(121 + 120) + 111

= 11.241 + 111

= 2651 + 111

= 2762