`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:

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

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

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

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

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

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

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

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

⇔\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

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

\(a,\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\\ =\dfrac{3\times4\times2\times15}{2\times5\times3\times4}\\ =\dfrac{3\times4\times2\times5\times3}{2\times5\times3\times4}\\ =3\)

\(b,\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\\ =\dfrac{6\times5\times7\times7\times8}{7\times8\times3\times6\times5}\\ =\dfrac{7}{3}\)

a) \(\dfrac{3}{2}\times\dfrac{4}{5}\times\dfrac{2}{3}\times\dfrac{15}{4}\)

\(=\left(\dfrac{3}{2}\times\dfrac{2}{3}\right)\times\left(\dfrac{4}{5}\times\dfrac{15}{4}\right)\)

\(=1\times3\)

\(=3\)

b) \(\dfrac{6}{7}\times\dfrac{5}{8}\times\dfrac{7}{3}\times\dfrac{7}{6}\times\dfrac{8}{5}\)

\(=\left(\dfrac{6}{7}\times\dfrac{7}{6}\right)\times\left(\dfrac{5}{8}\times\dfrac{8}{5}\right)\times\dfrac{7}{3}\)

\(=1\times1\times\dfrac{7}{3}\)

\(=\dfrac{7}{3}\)

Chúc bạn học tốt

Tổng trên có: (2002 -1):1+1 =2002 (số hạng)

Ta có: 1+ (2 -3 -4 +5)+ (6 -7 -8 +9)+...+(1998 -1999 -2000 -2001)+ 2002   (có 500 nhóm, thừa 2 số)

       = 1 + 0 + 0 + ... + 0 + 2002

       = 2003

Chúc bạn học tốt.

cac ban tra loi giup mk voi 

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

Số số hạng của dãy: (2012-2):1+1=2011(số)

Vì 2011 là số lẻ nên A là số âm.

a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)

\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)

\(A=27\cdot50+73\cdot50\)

\(A=50\cdot\left(27+73\right)\)

\(A=50\cdot100\)

\(A=5000\)

b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)

\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)

\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)

\(B=2^3\cdot5^3\)

\(B=10^3\)

\(B=1000\)

Cách 1 : A=100+98+96+...+2-97-95-...-1

A= 100 + (98-97) + (96-95) + ... +(2-1)

Từ 1 đến 98 có 98 số => có 98 : 2 cặp mà hiệu = 1

A = 100 + 49 x 1 = 149

B = 1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302

B = 1 + 2 + (302 - 300) + (301 - 299) + ... + (10 - 8) + (9-7) + (6-4) + (5-3) 

Từ 3 đến 302 có 300 số => có 300 : 2 cặp hiệu = 2

B = 1 + 2 + 150 x 2 = 303

Cách  2 :

A = 100 + (98-97) + (96-95) + ……. + (2-1)

Ta thấy:  97; 95; ….; 1  có (97 – 1) : 2 + 1 = 49 (số hạng)

A = 100 + (1+1+1+….+1)     (có 49 số 1).

A = 100 + 49 = 149

a, A = 100+(98-97)+(86-95)+....+(2-1) = 100+1+1+...+1 (49 số 1) = 149

b, B = 1+(2-3-4+5)+(6-7-8+9)+....(297-298-299+330)+331-332

= 1+0+0+....+0+331-332 = 0 

Nếu đúng thì k mk nha

A)6 và 4/5-(1 và 2/3+3 và 4/5)

=6+4/5-1-2/3-3-4/5

=(6-1-3)+(4/5-4/5)-2/3

=2-2/3=1 và 1/3

B)[(19+5/8/7/12)-(13+1/4/7/12)]X4/5

=[(19-13)+(5/8/7/12-1/4/7/12)]X4/5

=[6+3/8/7/12]X4/5

=[6+9/14]X4/5

=93/14 X4/5=186/35

nhớ

A=100+(98-97)+(96-95)+...+(2-1)

=100+1+1+1+...+1

Từ 2 đến 98 có

(98-2):2+1=49 (số hạng)

=>A=100+1.49

=100+49

=149