1)

a)2008.213+2008.87

=2008(213+87)

=2008.300

=602400

b)2007.75+25.2007

=2007(75+25)

=2007.100

=200700

c)28.76+28.13+11.28

=28(76+13+11)

=28.100

=2800

d)6^2:43+2.5^2

=36:43+2.25

=36/43+50

=36/43+2150/43

=2186/43

e)3^3.118-3^3.18

=27.118-27.18

=27(118-18)

=27.100

=2700

f)4.5^2-3.2^3

=4.25-3.8

=100-24

=76

g)15.2^3+4.3-5.7

=15.8+12-35

=120+12-35

=97

h)2^2.3-(1^10+8):3^2

=4.3-(1+8):9

=12-9:9

=12-1

=11

i)4^8:4^5-(1^30+17):3^2

=4^3-(1+17):9

=64-18:9

=64-2

=62

j)150-[10^2-(14-11)^2.2007^0]

=150-[100-3^2.1]

=150-[100-9]

=150-91

=59

k)12:{390:[500-(125+35.7)]}

=12:{390:[500-(125+245)]}

=12:{390:[500-370]}

=12:{390:130}

=12:3

=4

l)1+2+..+2013

SSH=(2013-1):1+1

       =2012:1+1

       =2013

Tổng=(2013+1).2013:2

         =2014.2013:2

         =2027091

m)21.7^2-11.7^2+90.7^2+49.125.16

=21.49-11.49+90.49+49.2000

=49(21-11+90+2000)

=49.2100

=102900

n)(2^2+2^1+2^2+2^3).2^0.2^1.2^2.2^3

=(4+2+4+8).1.2.4.8

=18.1.2.4.8

=36.4.8

=144.8

=1152

Bài 2:

a)B={x∈NI30<x≤40}

⇒B={31; 32; 33; 34; 35; 36; 37; 38; 39; 40}

b)4(3x-4)-2=18

12x-16-2=18

12x-18=18

12x=18+18

12x=0

x=0:12

⇒x∈{∅}

c)(105-x):2^5=3^0+1

(105-x):32=1+1

(105-x):32=2

(105-x)=2.32

(105-x)=64

x=105-64

x=41

d)2x-138=2^2.3^2

2x-138=4.9

2x-138=36

2x=36+138

2x=174

x=174:2

x=87

e)(6x-39)28=5628

(6x-39)=5628:28

(6x-39)=201

6x=201+39

6x=240

x=240:6

x=40

f)(9x+2)3=60

27x+6=60

27x=60-6

27x=54

x=54:27

x=2

g)(26-3x):5+71=75

(26-3x):5=75-71

(26-3x):5=4

(26-3x)=4.5

(26-3x)=20

3x=26-20

3x=3

x=3:3

x=1

h)5^x+1=125

5^x+1=5^3

x+1=3

x=3-1

x=2

Bài 3:

a)126÷x; 210÷x và 10<x<40

⇒x∈ƯC(126; 210)

Ta có:

126=2x3^2x7

210=2x3x5x7

⇒ƯCLN(210; 126)=2x3x7=42

⇒ƯC(126; 210)=Ư(42)={1; 2; 3; 6; 7; 14; 21; 42}

Mà 10<x<40, nên:

⇒x∈{14; 21}

b)x÷18; x÷15; x÷12 và 200≤x≤500

⇒x∈BC(18; 15; 12)

Ta có:

18=2x3^2

15=3x5

12=2^2x3

⇒BCNN(18; 15; 12)=2^2x3^2x5=180

⇒BC(18; 12; 15)=B(180)={0; 180; 360; ...}

Mà 200≤x≤500 nên:

⇒x∈{360}

c)x:2; x:3; x:4; x:5 thì đều dư 1 và 100<x<150

⇒(x-1)∈BC(2; 3; 4; 5)

Ta có:

2=2.1

3=1.3

4=2^2

5=5.1

⇒BCNN(2; 3; 4; 5)=2x3x4x5=120

⇒BC(2; 3; 4; 5)=B(120)={0; 120; 240; 360;...}

⇒x∈{1; 121; 241; 361;...}

Mà 100<x<150, nên:

x∈{121; 241; 361; 481}