Câu 5:

a: \(31\cdot\left(-18\right)+31\cdot\left(-81\right)-31\)

\(=31\left(-18-81-1\right)\)

\(=31\cdot\left(-100\right)=-3100\)

b: \(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)

\(=\left(-12\right)\left(47+52+1\right)\)

\(=-12\cdot100=-1200\)

c: \(13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\)

\(=13\cdot45-3\cdot45\)

\(=45\cdot10=450\)

d: \(-48+48\left(-78\right)+48\left(-21\right)\)

\(=48\left(-1-78-21\right)\)

\(=48\left(-100\right)=-4800\)

Câu 4:

a: \(\left(-6-2\right)\left(-6+2\right)=\left(-8\right)\cdot\left(-4\right)=32\)

b: \(\dfrac{\left(7\cdot3-3\right)}{-6}=\dfrac{21-3}{-6}=\dfrac{18}{-6}=-3\)

c: \(\left(-5+9\right)\cdot\left(-4\right)=4\cdot\left(-4\right)=-16\)

d: \(\dfrac{72}{-6\cdot2+4}=\dfrac{72}{-12+4}=\dfrac{72}{-8}=-9\)

a: =25x100-150=2500-150=2350

c: \(=520:\left\{515\cdot25\right\}\)

=104/2575

Bài 2 

a: =>x=-23-7=-30

b: =>75:x=15

hay x=5

c: =>2(x+4)=80

=>x+4=40

hay x=36

d: \(\Leftrightarrow2^x\cdot11=88\)

hay x=3

Bài 3: 

Có thể chia được nhiều nhất 12 tổ vì UCLN(180;132)=12

Khi đó, mỗi tổ có 15 nam và 11 nữ

`5`

`a, -7/21 +(1+1/3)`

`=-7/21 + ( 3/3 + 1/3)`

`=-7/21+ 4/3`

`=-7/21+ 28/21`

`= 21/21`

`=1`

`b, 2/15 + ( 5/9 + (-6)/9)`

`= 2/15 + (-1/9)`

`= 1/45`

`c, (9-1/5+3/12) +(-3/4)`

`= ( 45/5-1/5 + 3/12)+(-3/4)`

`= ( 44/5 + 3/12)+(-3/4)`

`= 9,05 +(-0,75)`

`=8,3`

`6`

`x+7/8 =13/12`

`=>x= 13/12 -7/8`

`=>x=5/24`

`-------`

`-(-6)/12 -x=9/48`

`=> 6/12 -x=9/48`

`=>x= 6/12-9/48`

`=>x=5/16`

`---------`

`x+4/6 =5/25 -(-7)/15`

`=>x+4/6 =1/5 + 7/15`

`=> x+ 4/6=10/15`

`=>x=10/15 -4/6`

`=>x=0`

`----------`

`x+4/5 = 6/20 -(-7)/3`

`=>x+4/5 = 6/20 +7/3`

`=>x+4/5 = 79/30`

`=>x=79/30 -4/5`

`=>x= 79/30-24/30`

`=>x= 55/30`

`=>x= 11/6`

\(5)\)

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=1\)

\(--------------\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)

\(B=\dfrac{1}{45}\)

\(------------\)

\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{83}{10}\)

\(6)\)

\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}-\dfrac{7}{8}\)

\(x=\dfrac{104}{96}-\dfrac{84}{96}\)

\(x=\dfrac{5}{24}\)

\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)

\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)

\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)

\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)

\(x=\dfrac{-11}{16}\)

\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)

\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)

\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)

\(x+\dfrac{4}{6}=\dfrac{12}{25}\)

\(x=\dfrac{12}{25}-\dfrac{4}{6}\)

\(x=\dfrac{72}{150}-\dfrac{100}{150}\)

\(x=\dfrac{-14}{75}\)

\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)

\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)

\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)

\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

\(x=\dfrac{79}{30}-\dfrac{4}{5}\)

\(x=\dfrac{79}{30}-\dfrac{24}{30}\)

\(x=\dfrac{11}{6}\)

Giải:

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=\dfrac{21}{21}=1\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{6}{45}+\dfrac{-5}{45}\)

\(B=\dfrac{1}{45}\)

\(C=\left(9-\dfrac{1}{5}+\dfrac{3}{12}\right)+\dfrac{-3}{4}\)

\(C=\left(\dfrac{44}{5}+\dfrac{1}{4}\right)+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{166}{20}\)

Bài 6:

\(a)x+\dfrac{7}{8}=\dfrac{13}{12}\)

⇔\(x=\dfrac{13}{12}-\dfrac{7}{8}=\dfrac{5}{24}\)

\(b)\dfrac{-6}{12}-x=\dfrac{9}{48}\)

⇔\(x=\dfrac{-6}{12}-\dfrac{9}{48}=\dfrac{-11}{16}\)

\(c)x+\dfrac{4}{6}=\dfrac{5}{25}-\dfrac{-7}{15}\)

\(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{2}{3}=0\)

\(d)x+\dfrac{4}{5}=\dfrac{6}{20}-\dfrac{-7}{3}\)

⇔\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

⇔\(x=\dfrac{79}{30}-\dfrac{4}{5}=\dfrac{11}{6}\)

a, 10 -12 - 8

= 10 - (12 + 8 )

= 10 - 20 = -10

b, 4-(-15)-5+6

= 4+15-5+6

= (4+6) + (15-5)

= 10+10=20

c, 2-12-4-6

= (2-12)-(4+6)

= (-10) - 10

=-20

d, -45-5-(-12)+8

= -45-5+12+8

= [(-45)-5]+(12+8)

= (-50)+20

= -30

\(a.10-12-8\\ =10+\left(-12-8\right)\\ =10+\left(-20\right)=-10\\ b.4-\left(-15\right)-5+6\\ =4+5-5+6\\ =\left(4+6\right)+\left(15-5\right)\\ =10+10=20\\ c.2-12-4-6\\ =\left(2-12\right)+\left(-4-6\right)\\ =-10+\left(-10\right)=-20\\ d.-45-5-\left(-12\right)+8\\ =-45-5+12+8\\ =\left(-45-5\right)+\left(12+8\right)\\ =-50+20=-30\)

Bài 1:

1: =15+37+52-37-17=52-2=50

2: =38-42+14-25+27+15=62-42+29=20+29=49

Bài 1: Bỏ ngoặc rồi tính

3) (21-32) - (-12+32)=21-32-(-12)-32=21-32+12-32=-31

4) (12+21) - (23-21+10)=12+21-23+21-10=21

5) (57-725) - (605-53)=57-725-605+53=-1220

6) (55+45+15) - (15-55+45)=55+45+15-15+55-45=55+55=110

Bài 2: Tính các tổng sau một cách hợp lí

1) (-37) + 14 + 26 + 37=(-37+37)+(14+26)=0+40=40

2) (-24) +6 + 10 + 24=(-24+24)+(6+10)=0+16=16

3) 15 + 23 + (-25) + (-23)=(15-25)+(23-23)=-10+0=-10

4) 60 + 33 + (-50) + (-33)=(60-50)+(33-33)=10+0=10

5) (-16) + (-209) + (-14) + 209=(-16-14)+(-209+209)=-30+0=-30

6) (-12) + (-13) + 36 + (-11)=(-11-12-13)+36=-36+36=0

=15+37+52-37-17 =(37-37)+52+15-17 =0+52+15-17 =67-17 =50

(38 – 42 + 14) – (25 – 27 – 15) =38-42+14-25+27+15 =27 c. -(21 – 32) – (-12 + 32) =-21+32+12-32 =-9 d. -(12 + 21 – 23) - (23 – 21 + 10) =-12-21+23-23+21-10 =-22 e. (57 – 725) – (605 – 53) =57-725-605+53 =-1220 f. (55 + 45 + 15) – (15 – 55 + 45) =55+45+15-15+55-45 =110

a) (-37) + 14 + 26 + 37

= [(-37) + 37] + (14 + 26)

= 0 + 40 = 40

b) (-24) + 6 + 10 + 24

= [(-24) + 24] + (10 + 6)

= 0 + 16 = 16

c) 15 + 23 + (-25) + (-23)

= [15 + (-25)] + [23 + (-23)]

= (-10) + 0 = -10

d) 60 + 33 + (-50) + (-33)

= [60 + (-50)] + [33 + (-33)]

= 10 + 0 = 10

e) (-16) + (-209) + (-14) + 209

= [(-16)  + (-14)] + [(-209) + 209]

= (-30) + 0 = -30

f) \(-3^2+\left(-54\right)\div\left[\left(-2\right)^8+7\right]\times\left(-2\right)^2\\ =\left(-9\right)+\left(-54\right)\div263\times4\\ =\left(-9\right)+\dfrac{-216}{263}=\dfrac{-2583}{263}\)

a. \(\left[\left(-37\right)+37\right]+\left(14+16\right)\) = 30 

B. \(\left[\left(-24\right)+24\right]+\left(10+6\right)\) = 16

C. \(\left[\left(-23\right)+23\right]+\left(15-23\right)\)= -8

d. \(\left[33-33\right]+\left(60-50\right)\) = 10

e. \(\left(209-209\right)+\left(-16-14\right)\)= -30

a) (44 x 52 x 60) : (11 x 13 x 15)

= (44 : 11) x ( 52 : 13) x (60 : 15)

= 4 x 4 x 4 = 64

d) 1 + 6 + 11 + 16 + ... + 46 + 51

Ta có : 1 + 6 + 11 + 16 + ... + 46 + 51 ( có 11 số )

= (51 + 1) x 11 : 2 = 286

a, Ta thấy 44:11=4               ,         52;13=4        ,                60:15=4,                                                                                                                         => 4*4*4=64             

b,Ta có : 16*17-5=16*(16+1)-5=16*16+16-5=16*16+11    Ta thấy Cụm 1 = Cụm 2 => Đáp án =1