\(\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}\)

b) 

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Mua vip
VT
Vũ Thị Thanh Thủy
2 tháng 3 2018

\(a,\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}\)

\(b,\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}\)

\(c,\dfrac{3}{13}-\dfrac{5}{-8}+\dfrac{-1}{2}\)

\(d,\dfrac{1}{2}+\dfrac{1}{-3}\)
#Hỏi cộng đồng OLM #Toán lớp 6
2
NT
Nguyen Thi Huyen
2 tháng 3 2018

\(a.\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}=\dfrac{12}{20}-\dfrac{-14}{20}-\dfrac{-13}{20}=\dfrac{12-\left(-14\right)-\left(-13\right)}{20}=\dfrac{39}{20}\)

\(b.\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}=\dfrac{3}{4}+\left(\dfrac{-6}{18}-\dfrac{5}{18}\right)=\dfrac{3}{4}+\dfrac{-11}{18}=\dfrac{27}{36}-\dfrac{-22}{36}=\dfrac{49}{36}\)

\(c.\dfrac{3}{13}-\dfrac{5}{-8}+\dfrac{-1}{2}=\dfrac{3}{13}-\left(\dfrac{5}{-8}+\dfrac{-4}{8}\right)=\dfrac{3}{13}-\dfrac{1}{8}=\dfrac{24}{104}-\dfrac{13}{104}=\dfrac{11}{104}\)

\(d.\dfrac{1}{2}+\dfrac{1}{-3}=\dfrac{3}{6}+\dfrac{-2}{6}=\dfrac{1}{6}\)

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VA
Việt Anh
2 tháng 3 2018

\(a,\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}\)

\(=\dfrac{12}{20}+\dfrac{14}{20}+\dfrac{13}{20}\)

\(=\dfrac{12+14+13}{20}\)

\(=\dfrac{39}{20}\)

\(b,\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}\)

\(=\dfrac{27}{36}+\dfrac{-12}{36}-\dfrac{10}{36}\)

\(=\dfrac{27+\left(-12\right)-10}{36}\)

\(=\dfrac{5}{36}\)

\(c,\dfrac{3}{13}-\dfrac{5}{-8}+\dfrac{-1}{2}\)

\(=\dfrac{24}{104}-\dfrac{-65}{104}+\dfrac{-52}{104}\)

\(=\dfrac{24-\left(-65\right)+\left(-52\right)}{104}\)

\(=\dfrac{37}{104}\)

\(d,\dfrac{1}{2}+\dfrac{1}{-3}\)

\(=\dfrac{3}{6}+\dfrac{-2}{6}\)

\(=\dfrac{3+\left(-2\right)}{6}\)

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

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SG
Sách Giáo Khoa
17 tháng 4 2017

Tính :

a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)

b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)

c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)

d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)
#Hỏi cộng đồng OLM #Toán lớp 6
2
Q
Quìn
17 tháng 4 2017

a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)

\(=\dfrac{17}{24}-\dfrac{7}{12}\)

\(=\dfrac{1}{8}\)

b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)

\(=\dfrac{23}{56}-\dfrac{1}{2}\)

\(=\dfrac{-5}{56}\)

c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)

\(=\dfrac{-5}{12}-\dfrac{11}{18}\)

\(=\dfrac{-37}{36}\)

d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)

\(=\dfrac{2}{3}-\dfrac{1}{13}-\dfrac{7}{8}\)

\(=\dfrac{23}{39}-\dfrac{7}{8}\)

\(=\dfrac{-89}{312}\)

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NT
nguyễn thị thúy
18 tháng 4 2017

2016-04-06_091251

2016-04-06_091238

2016-04-06_091259

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D
dangthuylinh
23 tháng 7 2017
a)\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{7}_{ }+\dfrac{5}{-8}\) b)\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}_{...
Đọc tiếp

a)\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{7}_{ }+\dfrac{5}{-8}\)

b)\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}_{ }+\dfrac{-8}{13}\)

c)\(\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

d)\(\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

e)\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

f)\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
#Hỏi cộng đồng OLM #Toán lớp 6
1
D
dangthuylinh
23 tháng 7 2017

Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OKleuleu

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LG
Lê Gia Bảo
23 tháng 7 2017

Mình khuyen bạn phải suy nghĩ kĩ bài trước khi đăng lên nhé!!hum

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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)...
Đọc tiếp

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) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}\dfrac{1}{5}\right)\)
d)\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
e)\(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
f)\(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
h)\(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
i)\(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
k)\(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
Bài 2: Tính bằng cánh hợp lí
A=\(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
B=\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
C=\(\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
D=\(10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)
#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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

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

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

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

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PN
Phạm Ngân Hà
29 tháng 4 2017

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

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

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

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

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

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

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

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

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

\(B=17+4,03\)

\(B=21,03\)

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

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

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

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

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

\(C=75\)

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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,\(\dfrac{3}{4}\):...
Đọc tiếp

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,\(\dfrac{3}{4}\): x =\(\dfrac{-7}{12}\)

d,\(\dfrac{3}{8}\)-\(\dfrac{1}{6}\)- x = \(\dfrac{1}{4}\)

e, \(\dfrac{1}{3}\)+\(\dfrac{1}{2}\): x = -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

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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 vuiko có tâm trạng mà cười nữalolanglimdim

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NV
Nguyễn Vũ Hoàng
27 tháng 11 2017
Tính giá trị biểu thức : 1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\) 2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\) 3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\) 4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\) 5....
Đọc tiếp

Tính giá trị biểu thức :

1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\)

2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)

4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)

5. Cho \(M=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\) ; \(N=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\). Tính \(P=M-N\)

6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)

7. \(F=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{256}+\dfrac{3}{64}}{1-\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)

9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)

10. \(I=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)

11. \(K=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{999}\right)\)

12. \(L=1\dfrac{1}{3}+1\dfrac{1}{8}+1\dfrac{1}{15}...\) (98 thừa số)

13. \(M=-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{3}}}}\)

14. \(N=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}\)

15. \(P=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{5}-1\right)...\left(\dfrac{1}{2001}-1\right)\)

16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)
#Hỏi cộng đồng OLM #Toán lớp 6
2
QH
Quang Ho Si
27 tháng 11 2017

1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)

2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)

3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)

4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

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NA
Nguyễn Anh Quang
29 tháng 4 2022

hôi lì sít

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KK
Kirigaya Kazuto
27 tháng 2 2017
So sánh A và B : a) \(A=\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}\) \(B=\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\) b) \(A=\dfrac{3}{8^3}+\dfrac{7}{8^4}\) \(B=\dfrac{7}{8^3}+\dfrac{3}{8^4}\) c) \(A=\dfrac{10^7+5}{10^7-8}\) \(B=\dfrac{10^8+6}{10^8-7}\) d) \(A=\dfrac{10^{1992}+1}{10^{1991}+1}\) \(B=\dfrac{10^{1993}+1}{10^{1992}+1}\) ...
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So sánh A và B :

a) \(A=\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}\)

\(B=\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)

b) \(A=\dfrac{3}{8^3}+\dfrac{7}{8^4}\)

\(B=\dfrac{7}{8^3}+\dfrac{3}{8^4}\)

c) \(A=\dfrac{10^7+5}{10^7-8}\)

\(B=\dfrac{10^8+6}{10^8-7}\)

d) \(A=\dfrac{10^{1992}+1}{10^{1991}+1}\)

\(B=\dfrac{10^{1993}+1}{10^{1992}+1}\)
#Hỏi cộng đồng OLM #Toán lớp 6
2
NQ
Nguyễn Quang Ngọc Trác
20 tháng 3 2017

d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B

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TL
Trèo lên cột điện thế hiện tình yêu
26 tháng 3 2017

cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A

Suy ra B>A(chuc ban hoc goi nhe)

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VM
van mai huong
21 tháng 3 2017
1.Tính a) (\(\dfrac{-2}{3}\)+1\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)).\(\dfrac{-24}{10}\) b) \(\dfrac{13}{15}\).0,25.3+(\(\dfrac{8}{15}\)-\(1\dfrac{19}{60}\)):\(1\dfrac{23}{24}\) c) (\(\dfrac{12}{32}\)+\(\dfrac{5}{-20}\)-\(\dfrac{10}{24}\)):\(\dfrac{2}{3}\) d) \(4\dfrac{1}{2}\):(2,5-\(3\dfrac{3}{4}\))+(\(\dfrac{-1}{2}\))\(^2\) e) \(\dfrac{-5}{2}\):(\(\dfrac{3}{4}\)-\(\dfrac{1}{2}\)) 2. Tính nhanh a)...
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1.Tính

a) (\(\dfrac{-2}{3}\)+1\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)).\(\dfrac{-24}{10}\)

b) \(\dfrac{13}{15}\).0,25.3+(\(\dfrac{8}{15}\)-\(1\dfrac{19}{60}\)):\(1\dfrac{23}{24}\)

c) (\(\dfrac{12}{32}\)+\(\dfrac{5}{-20}\)-\(\dfrac{10}{24}\)):\(\dfrac{2}{3}\)

d) \(4\dfrac{1}{2}\):(2,5-\(3\dfrac{3}{4}\))+(\(\dfrac{-1}{2}\))\(^2\)

e) \(\dfrac{-5}{2}\):(\(\dfrac{3}{4}\)-\(\dfrac{1}{2}\))

2. Tính nhanh

a) \(\dfrac{5}{23}\).\(\dfrac{17}{26}\)+\(\dfrac{5}{23}\).\(\dfrac{10}{26}\)-\(\dfrac{5}{23}\)

b) \(\dfrac{1}{7}\).\(\dfrac{5}{9}\)+\(\dfrac{5}{9}\).\(\dfrac{2}{7}\)+\(\dfrac{5}{9}\).\(\dfrac{1}{7}\)+\(\dfrac{5}{9}\).\(\dfrac{3}{7}\)
#Hỏi cộng đồng OLM #Toán lớp 6
2
NL
Nanami Luchia
21 tháng 3 2017

2) Tinh nhanh:

a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)

= \(\dfrac{5}{598}\)

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NL
Nanami Luchia
21 tháng 3 2017

b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)

= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)

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NG
Nguyễn Gia Hân
27 tháng 7 2017
a)(\(\dfrac{2}{9}\)(6x-\(\dfrac{3}{4}\))-3(\(\dfrac{1}{4}\)x-\(\dfrac{1}{5}\))=\(\dfrac{-8}{15}\) b)\(\dfrac{3}{4}\)(x-\(\dfrac{8}{9}\))-5(x-\(\dfrac{11}{15}\))+\(\dfrac{1}{2}\)x=5\(\dfrac{1}{12}\) c)\(\dfrac{2}{3}\)(-3x-\(\dfrac{6}{5}\))-\(\dfrac{5}{7}\)(\(\dfrac{49}{15}\)-14x)=-2\(\dfrac{1}{3}\) d)(12\(\dfrac{7}{8}\)-10\(\dfrac{13}{8}\)):x-1\(\dfrac{7}{33}\):\(\dfrac{8}{11}\)=1\(\dfrac{2}{3}\) ...
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a)(\(\dfrac{2}{9}\)(6x-\(\dfrac{3}{4}\))-3(\(\dfrac{1}{4}\)x-\(\dfrac{1}{5}\))=\(\dfrac{-8}{15}\)

b)\(\dfrac{3}{4}\)(x-\(\dfrac{8}{9}\))-5(x-\(\dfrac{11}{15}\))+\(\dfrac{1}{2}\)x=5\(\dfrac{1}{12}\)

c)\(\dfrac{2}{3}\)(-3x-\(\dfrac{6}{5}\))-\(\dfrac{5}{7}\)(\(\dfrac{49}{15}\)-14x)=-2\(\dfrac{1}{3}\)

d)(12\(\dfrac{7}{8}\)-10\(\dfrac{13}{8}\)):x-1\(\dfrac{7}{33}\):\(\dfrac{8}{11}\)=1\(\dfrac{2}{3}\)
#Hỏi cộng đồng OLM #Toán lớp 6
5
NG
Nguyễn Gia Hân
27 tháng 7 2017

Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya

e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-22).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))

f)| 5x+21 | = | 2x -63 |

g) -45 - |-3x-96 | - 54=-207

Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)

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KL
Khánh Linh
1 tháng 8 2017

a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)

<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)

<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)

<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)

<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân

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