1. so sánh A với 2, biết A = \(\dfrac{1}{2}\)+\(\dfrac{2}{2^2}\)

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

Khách

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Hoàng Lê Kim Ngân

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10 tháng 4 2024 - olm

1. so sánh A với 2, biết A = \(\dfrac{1}{2}\)+\(\dfrac{2}{2^2}\)+\(\dfrac{3}{2^3}\)+...+\(\dfrac{2023}{2^{2023}}\).

2.so sánh P =\(\dfrac{10^7+5}{10^7-8}\) và Q = \(\dfrac{10^8+6}{10^8-7}\).

3.cho A = 1- \(\dfrac{3}{4}\) +(\(\dfrac{3}{4}\))\(^2\) - (\(\dfrac{3}{4}\))\(^3\) + (\(\dfrac{3}{4}\))\(^4\) - ...-(\(\dfrac{3}{4}\))\(^{2023}\).

chứng minh A không là số nguyên.

4. tính C = \(\dfrac{32}{3.7}\)+\(\dfrac{6}{7.41}\)+\(\dfrac{9}{41.10}\)+\(\dfrac{1}{10.51}\)+\(\dfrac{19}{51.14}\).

#Hỏi cộng đồng OLM #Toán lớp 6

Nguyễn Thị Thương Hoài

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11 tháng 4 2024

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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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#Hỏi cộng đồng OLM #Toán lớp 6

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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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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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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#Hỏi cộng đồng OLM #Toán lớp 6

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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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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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,\(\dfrac{3}{4}\):...

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#Hỏi cộng đồng OLM #Toán lớp 6

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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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

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hello hello

15 tháng 4 2018

1. tính a. \(\dfrac{3}{5}\)-\(\dfrac{2}{3}\)+\(\dfrac{16}{15}\) b. (\(\dfrac{5}{6}\)-\(\dfrac{4}{5}\)):\(\dfrac{7}{30}\)-\(\dfrac{6}{7}\) c....

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#Hỏi cộng đồng OLM #Toán lớp 6

thảo nguyễn

17 tháng 4 2018

Violympic toán 6

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Phạm Thu Huyền

20 tháng 3 2017

Bài 1 Tính a) ( \(\dfrac{-2}{3}\) + \(1\dfrac{1}{4}\) - \(\dfrac{1}{6}\) ) : \(\dfrac{-24}{10}\) b) \(\dfrac{13}{15}\) x 0.25 x 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 - \(3\dfrac{3}{4}\) ) + ( -\(\dfrac{1}{2}\) ) e) \(\dfrac{-5}{2}\) : ( \(\dfrac{3}{4}\) - \(\dfrac{1}{2}\)...

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#Hỏi cộng đồng OLM #Toán lớp 6

Nguyễn Đắc Định

20 tháng 3 2017

a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)

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

=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)

=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)

b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)

=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)

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

=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)

= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)

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

=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)

=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)

e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)

=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)

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ngô trọng tấn

7 tháng 3 2017

...

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#Hỏi cộng đồng OLM #Toán lớp 6

Tiểu Long Nữ Thánh Hiền

7 tháng 3 2017

CÁCH 1 : A = \(\dfrac{235}{11}-\left(\dfrac{8}{5}+\dfrac{81}{11}\right)\)

A = \(\dfrac{235}{11}-\left(\dfrac{88}{55}+\dfrac{405}{55}\right)\)

A = \(\dfrac{235}{11}-\dfrac{493}{55}\)

A = \(\dfrac{1175}{55}+\dfrac{493}{55}\)

A = \(\dfrac{1668}{55}\)

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dangthuylinh

22 tháng 3 2017

Bài 2 : Tính a) \(\dfrac{3}{4}\) - \(\dfrac{-5}{2}\) - \(\dfrac{7}{-24}\) b) \(\dfrac{4}{7}\) + \(\dfrac{-5}{8}\) - \(\dfrac{3}{28}\) c) \(\dfrac{7}{36}\) - \(\dfrac{8}{-9}\) + \(\dfrac{-2}{3}\) d) \(\dfrac{-1}{2}\) + \(\dfrac{3}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{-7}{18}\) + \(\dfrac{4}{7}\) e) \(\dfrac{2}{7}\) + \(\dfrac{-3}{8}\) + \(\dfrac{11}{7}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{7}\) +...

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#Hỏi cộng đồng OLM #Toán lớp 6

Doraemon

22 tháng 3 2017

\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{13}{4}-\dfrac{7}{-24}\)

\(=\dfrac{85}{24}\)

\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{-3}{56}-\dfrac{3}{28}\)

\(=\dfrac{-9}{56}\)

\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)

\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)

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

\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-4}{7}+\dfrac{4}{7}\)

\(=0\)

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

\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{47}{24}+\dfrac{5}{-8}\)

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

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Quìn

22 tháng 3 2017

Bài 2 : Tính

a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)

\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)

\(=\dfrac{85}{24}\)

b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)

\(=\dfrac{32+\left(-35\right)-6}{56}\)

\(=\dfrac{-9}{56}\)

c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)

\(=\dfrac{7-32+\left(-24\right)}{36}\)

\(=\dfrac{-49}{36}\)

d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\left(-1\right)+1\)

\(=0\)

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

\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)

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

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

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

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đỗ hương giang

9 tháng 4 2018

Tính ( bằng cách thuận tiên nếu có thể ) a, A = \(\dfrac{3^2}{10}\)+\(\dfrac{3^2}{40}\)+\(\dfrac{3^2}{8^8}\)+...+\(\dfrac{3^2}{340}\) b, A = \(\dfrac{5^2}{1\cdot6}\)+\(\dfrac{5^2}{6\cdot11}\)+...+\(\dfrac{5^2}{26\cdot31}\) c, A = \(\dfrac{1}{2}\)+\(\dfrac{2}{2\cdot4}\)+\(\dfrac{3}{4\cdot7}\)+\(\dfrac{4}{7\cdot11}\)+\(\dfrac{5}{11\cdot16}\) d, A = \(\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{249}+\dfrac{8}{609}\) e, A =...

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#Hỏi cộng đồng OLM #Toán lớp 6

Nguyễn Lê Phước Thịnh

29 tháng 6 2022

b: \(A=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)

=1-1/16=15/16

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‍‍‍‍‍‌‌‌‎‎

13 tháng 3 2018 - olm

4, so sánh A và B:

a,A=\(\dfrac{3}{8^3}+\dfrac{7}{8^4}\);B=\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)

b,A=\(\dfrac{10^7+5}{10^7-8}\);B=\(\dfrac{10^8+6}{10^8-7}\)

c,A=\(\dfrac{10^{1992}+1}{10^{1991}+1}\);B=\(\dfrac{10^{1993}+1}{10^{1992}+1}\)

#Hỏi cộng đồng OLM #Toán lớp 6

thai kim ngan

13 tháng 3 2018

a,A<B

b,A,<B

c,A<B

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13 tháng 3 2018

a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)

Vậy A < B

b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)

c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:

\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

Vậy A < B

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