1) âm năm phần 12

2) âm mười bảy phần 9

3) -1 

Đây là đáp án còn làm bài từ làm nhé

2:

a: =>2/3:x=1,4-2,4=-1

=>x=-2/3

b: =>x/5=25/30-19/30=6/30=1/5

=>x=1

3:

Số học sinh giỏi là 40*1/4=10 bạn

Số học sinh khá là 30*3/5=18 bạn

Số học sinh TB là 30-18=12 bạn

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

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

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

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

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

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

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}.\left(-\dfrac{9}{2}\right)=-\dfrac{3}{2}\)

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

a. 5/6

b. -6/7

c. 2/3

d. 2/9

e. -63/325

g. 43/20

a. 1/12 + 3/4

= 1/12+ 9/12

= 10/12 = 5/6

b. -4/7. 1 1/2

= -4/7. 3/2

= -6/7

c. 7/9+ ( 2/3 + -7/9)

= 7/9 + -1/9

= 8/9 = 2/3

d. 2/3 - 1/3 : 3/4

= 2/3 - 4/9

= 6/9 - 4/9

= 2/9

e. -7/25 . 11/13+ - 7/25. 2/13

=-77/ 325+ -14/ 325

= - 7/25

g. 2 2/5 . 0,25- ( 11/20+ 75%) : 13/5

= 12/5 . 1/4 - ( 11/20 + 3/4) : 13/5

= 12/5 . 1/4 - 13/10 : 12/ 5

= 3/5 - 13/24

= 7/120

lm đại vậy thôi nha hơi lâu vì mik bị lag nên phải làm lại

a) Ta có: \(\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-\dfrac{19}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\dfrac{-9}{16}\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}\cdot\dfrac{-16}{9}\)

\(=\dfrac{-112}{144}=\dfrac{-7}{9}\)

b) Ta có: \(17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)

\(=17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}\cdot\dfrac{4}{27}+350\%\)

\(=\dfrac{4}{27}\left(17+\dfrac{6}{11}-8-\dfrac{6}{11}\right)+\dfrac{7}{2}\)

\(=\dfrac{4}{27}\cdot9+\dfrac{7}{2}\)

\(=\dfrac{4}{3}+\dfrac{7}{2}=\dfrac{8}{6}+\dfrac{21}{6}=\dfrac{29}{6}\)

a: =11+3/4-6-5/6+4+1/2+1+2/3

=10+9/12-10/12+6/12+8/12

=10+13/12=133/12

b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)

=3-11/15

=34/15

c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)

\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)

d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^3+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^6.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}\left(3^5-3^4\right)}{2^6\left(3^6-3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9\left(7^3+2^3.7^3\right)}\)

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3+2^3.7^3}\) ( \(\frac{5\left(7^3-7^4\right)}{7^3.1+2^3.7^3}=\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\))

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\)

\(A=\frac{2^6.162}{486}-\frac{5.\left(-2058\right)}{7^3.9}\)

\(\Rightarrow A=\frac{648}{486}-\frac{-12540}{3087}\)

bạn tự tính nha máy tính mình mất r

bài 3:

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)

Bài 1:

a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)

\(=\dfrac{30}{30}-1\)

=1-1

=0

b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)

\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)

\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)

=3

c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)

\(=9-12\cdot\dfrac{9-8}{12}\)

=9-1

=8