D=\(2\dfrac{1}{3}-30\%+0,125-6,5\)

\(2\dfrac{1}{3}-30\%+0,125-6,5\)

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K

Khách

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NV

Nguyễn Văn Vi Duy Hưng

4 tháng 3 2023 - olm

D=\(2\dfrac{1}{3}-30\%+0,125-6,5\)

#Hỏi cộng đồng OLM #Toán lớp 6

1

LD

Long Duy

6 tháng 3 2023

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Các câu hỏi dưới đây có thể giống với câu hỏi trên

DP

Duong Pham

25 tháng 4 2018

Thực hiện phép tính(Tính nhanh nếu có thể): a) 0,25.\(\dfrac{12}{5}\).\(\dfrac{100}{7}\).49% ...

#Hỏi cộng đồng OLM #Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

3 tháng 7 2022

a: \(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{100}{7}\cdot\dfrac{49}{100}\)

\(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{49}{7}=\dfrac{3}{5}\cdot7=\dfrac{21}{5}\)

b: \(=\dfrac{3}{8}+\dfrac{1}{8}\cdot\dfrac{3}{4}-\dfrac{5}{4}\)

\(=\dfrac{12}{32}+\dfrac{3}{32}-\dfrac{40}{32}=\dfrac{-25}{32}\)

c: \(=\dfrac{4}{9}\left(\dfrac{-13}{27}-\dfrac{14}{27}\right)-\dfrac{5}{9}=\dfrac{-4}{9}-\dfrac{5}{9}=-1\)

d: \(=\dfrac{2}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{91\cdot95}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{91}-\dfrac{1}{95}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{92}{285}=\dfrac{46}{285}\)

VT

vũ tường vi

31 tháng 3 2017

bài 1 : tính a,\(\dfrac{-6}{11}\): (\(\dfrac{3}{5}.\)\(\dfrac{4}{11}\)) b, \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\) : 6 - \(\dfrac{11}{36}\) c, ( \(\dfrac{4}{5}\) +\(\dfrac{1}{2}\) ) : ( \(\dfrac{3}{13}\)-\(\dfrac{8}{13}\)) d, ( \(\dfrac{2}{3}\) -\(\dfrac{1}{4}\) +\(\dfrac{5}{11}\) ) : ( \(\dfrac{8}{12}\) +1 - \(\dfrac{7}{11}\) ) bài 2 , tìm x, biết a, \(\dfrac{-2}{5}\). x +\(\dfrac{4}{3}\) = \(\dfrac{7}{3}\) b,\(\dfrac{9}{8}\)+\(\dfrac{3}{8}\)-x = 0,125 c,\(\dfrac{3}{4}\):...

#Hỏi cộng đồng OLM #Toán lớp 6

7

DH

Đừng Hỏi Tên Tôi

31 tháng 3 2017

a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

VT

vũ tường vi

31 tháng 3 2017

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa

K

Kfkfj

10 tháng 7 2017

tính: a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\) b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\) c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\) d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\) e)...

#Hỏi cộng đồng OLM #Toán lớp 6

2

TV

Trai Vô Đối

10 tháng 7 2017

Cái này bn lầy máy tính ra tính tí là xong thôi

N

Nhok___Ngốc___Nghếch

5 tháng 5 2018

a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)

= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)

= \(2,75.\left(-3,2\right)\)

= \(-8,8\)

b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)

= \(\dfrac{3}{7}-\dfrac{2}{3}\)

= \(-\dfrac{5}{21}\)

c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)

= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)

= \(-\dfrac{23}{20}\)

d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)

= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)

=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)

= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)

= \(-\dfrac{23}{40}\)

e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)

= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)

= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)

= \(-\dfrac{5501}{225}\)

NT

Nguyễn Thu Thảo

15 tháng 3 2017

Tìm x, biết : a) \(\left(\dfrac{31}{20}-\dfrac{26}{45}\right).\dfrac{-36}{35}< x< \left(\dfrac{51}{56}+\dfrac{8}{21}+\dfrac{1}{3}\right).\dfrac{8}{13}\) b) \(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\) c) \(\left(\dfrac{1}{7}-\dfrac{1}{3}\right).x=\dfrac{28}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}\right)\) d) \(\dfrac{5}{7}.x=\dfrac{9}{8}-0,125\) ...

#Hỏi cộng đồng OLM #Toán lớp 6

2

HT

Hoàng Thị Ngọc Anh

15 tháng 3 2017

Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.

KW

Karry Wang

20 tháng 3 2017

b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)

=>\(\dfrac{-3}{5}.x=1\)

=>\(x=1:\dfrac{-3}{5}\)

=>\(x=\dfrac{-5}{3}\)

Vậy \(x=\dfrac{-5}{3}\)

PK

Phạm Kiều Trang

24 tháng 7 2017

Tính a, \(1\dfrac{7}{20}\) : 2,7+2,7:1,35+(0,4:\(2\dfrac{1}{2}\) ) . (4,2-\(1\dfrac{3}{40}\) ) b,...

#Hỏi cộng đồng OLM #Toán lớp 6

3

PN

Phạm Ngân Hà

14 tháng 10 2017

a) \(1\dfrac{7}{20}:2,7+2,7:1,35+\left(0,4:2\dfrac{1}{2}\right).\left(4,2-1\dfrac{3}{40}\right)\)

\(=\dfrac{27}{20}:\dfrac{27}{10}+\dfrac{27}{10}:\dfrac{27}{10}+\left(\dfrac{2}{5}:\dfrac{5}{2}\right).\left(\dfrac{21}{5}-\dfrac{43}{40}\right)\)

\(=1+1+\dfrac{4}{25}.\dfrac{25}{8}\)

\(=2+\dfrac{1}{2}\)

\(=2\dfrac{1}{2}\)

PN

Phạm Ngân Hà

14 tháng 10 2017

b) \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)

\(=\left(\dfrac{33}{5}-\dfrac{45}{14}\right).\dfrac{35}{6}:\left(-4\right):2,5\)

\(=\left(\dfrac{462}{60}-\dfrac{225}{60}\right).\dfrac{35}{6}.\dfrac{1}{-4}:\dfrac{5}{2}\)

\(=\dfrac{237}{60}.\dfrac{35}{6}.\dfrac{1}{-4}.\dfrac{2}{5}\)

\(=\dfrac{3.79.7.5.2}{5.14.3.2.\left(-4\right).5}\)

\(=\dfrac{79.7}{14.\left(-4\right).5}=\dfrac{553}{-280}\) (số xấu :v)

LQ

Liên Quân Mobile

12 tháng 8 2017

Tính nhanh .

a, \(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)

b, \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

#Hỏi cộng đồng OLM #Toán lớp 6

2

PT

Phạm Tú Uyên

12 tháng 8 2017

a) Hình như nhầm đề thì phải :v

\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)

\(=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{5}{11}}=1\)

b) \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}\right)}=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}\)

\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)

AB

Adonis Baldric

12 tháng 8 2017

a,\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}=\dfrac{\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right).132}{\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right).132}=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

b, Ta có : 0,125 = \(\dfrac{1}{8}\) ; 0,375 = \(\dfrac{3}{8}\) ; 0,2 = \(\dfrac{1}{5}\) ; 0,5 = \(\dfrac{3}{6}\)

\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)

\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)

\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)

LX

Linh Xinh

19 tháng 8 2017

tính a, A = \(\dfrac{3}{2}\) + \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)+ \(\dfrac{3}{250}\) + \(\dfrac{3}{1250}\) b, B = \(\dfrac{3}{2.5}\)+ \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + \(\dfrac{3}{11.14}\) c, C = \(\dfrac{5}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + \(\dfrac{5}{30}\) + \(\dfrac{5}{42}\) + ........... + \(\dfrac{5}{110}\) d, D = \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) + \(\dfrac{1}{4.5.6}\) + \(\dfrac{1}{5.6.7}\) + \(\dfrac{1}{6.7.8}\) +...

#Hỏi cộng đồng OLM #Toán lớp 6

1

PT

Phương Trâm

19 tháng 8 2017

Đăng ít thôi.

d) \(D=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+\dfrac{1}{7.8.9}+\dfrac{1}{8.9.10}\)

\(\Rightarrow2D=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}+\dfrac{2}{7.8.9}+\dfrac{2}{8.9.10}\)

\(\Rightarrow2D=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{4.5}-\dfrac{1}{5.6}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)

\(\Rightarrow2D=\dfrac{1}{2.3}-\dfrac{1}{9.10}\)

\(\Rightarrow2D=\dfrac{22}{45}\)

\(\Rightarrow D=\dfrac{11}{45}\)

HT

Huy Thắng Nguyễn

26 tháng 8 2017

Trả lời ít thôi.

T IÊU M

bà cha m ra :v

HM

Hoàng Mai Trang

2 tháng 5 2017

Tìm \(x,y\in Z\) biết

a) \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

b)\(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

c)\(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)

d)\(xy+2x-y=15\)

#Hỏi cộng đồng OLM #Toán lớp 6

1

DD

Dinh Dung

3 tháng 5 2017

các ý a,b,c c chỉ cần nhan chéo cho nhau

NN

no no

29 tháng 3 2017

Bài 1: Thực hiện phép tính a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\) b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\) c)...

#Hỏi cộng đồng OLM #Toán lớp 6

2

PN

Phạm Ngân Hà

28 tháng 4 2017

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

\(=\dfrac{1}{6}-\dfrac{7}{2}\)

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

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

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

\(=\dfrac{2}{2}=1\)

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

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

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

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

PN

Phạm Ngân Hà

29 tháng 4 2017

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)