VÒNG 2Bài 1: Mèo con nhanh nhẹn\(\dfrac{1}{2}\) \(\dfrac{1}{12}\)2 \(\dfrac{1}{6}\)\(\dfrac{1}{20}\)1 - \(\dfrac{1}{9}\) \(\dfrac{1}{15}\) \(\dfrac{2}{15}\)\(\dfrac{1}{2}\) \(\dfrac{2}{3}\)\(\...

SC

sữa cute

16 tháng 10 2021

VÒNG 2Bài 1: Mèo con nhanh nhẹn\(\dfrac{1}{2}\) + \(\dfrac{1}{12}\)2 + \(\dfrac{1}{6}\)\(\dfrac{1}{20}\)1 - \(\dfrac{1}{9}\) \(\dfrac{1}{15}\) + \(\dfrac{2}{15}\)\(\dfrac{1}{2}\) + \(\dfrac{2}{3}\)\(\dfrac{7}{12}\)\(\dfrac{4}{12}\) \(\dfrac{9}{14}\)+ \(\dfrac{1}{14}\)1 + \(\dfrac{1}{6}\)\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)\(\dfrac{1}{3}\) - \(\dfrac{2}{9}\) \(\dfrac{3}{2}\) + \(\dfrac{2}{3}\)\(\dfrac{1}{5}\)1 - \(\dfrac{8}{9}\) \(\dfrac{5}{7}\)1...

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#Toán lớp 5

0

NK

Nguyễn Kiều Anh

18 tháng 9 2021

Tính nhanh:

a, \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

b, \(\left(-\dfrac{1}{4}+\dfrac{7}{35}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

18 tháng 9 2021

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

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NK

Nguyễn Kiều Anh

18 tháng 9 2021

câu b, đâu ạ?

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H

Hà2k?

6 tháng 3 2023

\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\)

\(\dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\)

\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\)

\(\dfrac{1}{6}+\dfrac{2}{3}\) x \(\dfrac{8}{9}\)

#Toán lớp 4

1

C

Cihce

6 tháng 3 2023

\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\\ =\dfrac{3\times6}{5\times6}+\dfrac{1\times15}{2\times15}+\dfrac{8\times2}{15\times2}\\ =\dfrac{18}{30}+\dfrac{15}{30}+\dfrac{16}{30}\\ =\dfrac{49}{30}\\ \dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\\ =\dfrac{6\times2}{9\times2}+\dfrac{14}{18}-\dfrac{5\times3}{6\times3}\\ =\dfrac{12}{18}+\dfrac{14}{18}-\dfrac{15}{18}\\ =\dfrac{11}{18}\)

\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\\ =\dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}\\ =\dfrac{9}{20}-\dfrac{3}{20}\\ =\dfrac{6}{20}\\ =\dfrac{3}{10}\)

\(\dfrac{1}{6}+\dfrac{2}{3}\times\dfrac{8}{9}\\=\dfrac{1}{6}+\dfrac{16}{27}\\ =\dfrac{1\times9}{6\times9}+\dfrac{16\times2}{27\times2}\\ =\dfrac{9}{54}+\dfrac{32}{54}\\ =\dfrac{41}{54}.\)

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PN

Phương Nguyễn

25 tháng 1 2018

23) \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)

24) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

25) \(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 6 2022

24:

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)

\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)

=>(x+10)(x-2)=0

=>x=-10 hoặc x=2

25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

\(\Leftrightarrow x+5=0\)

hay x=-5

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KL

Khánh Linh

13 tháng 4 2017

BT1: Tính nhanh

1) \(\left(\dfrac{-4}{9}+\dfrac{3}{7}\right):1\dfrac{1}{15}+\left(\dfrac{4}{7}-\dfrac{5}{9}\right):1\dfrac{1}{15}\)

2) \(3\dfrac{2}{9}.15\dfrac{4}{7}-3\dfrac{2}{9}.8\dfrac{1}{7}+3\dfrac{2}{9}.\dfrac{15}{7}-3\dfrac{2}{9}.\dfrac{1}{7}\)

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 5 2022

1: \(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}\right)+\dfrac{16}{15}\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)

\(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}+\dfrac{4}{7}-\dfrac{5}{9}\right)=0\)

2: \(=\dfrac{29}{9}\left(15+\dfrac{4}{7}-8-\dfrac{1}{7}+\dfrac{15}{7}-\dfrac{1}{7}\right)\)

\(=\dfrac{20}{9}\cdot\left(7\cdot\dfrac{18}{7}\right)=\dfrac{20}{9}\cdot18=40\)

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HK

Hoàng Kim Nhung

8 tháng 7 2023

tính thuận tiện:\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\) \(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\) \(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\) \(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)mik sẽ chỉ tick 3 bn xong trước phải chi tiết rõ...

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#Toán lớp 4

2

NL

Nguyễn Lê Phước Thịnh

8 tháng 7 2023

a: =4/5+1/5+2/3+1/3=1+1=2

b: =17/12+7/12+29/7-8/7=3+2=5

c: =3/5+2/5+16/7-1/7-1/7

=1+2=3

d: =2/5+3/5+2/3+1/3+7/4+1/4

=1+1+2

=4

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HK

Hoàng Kim Nhung

8 tháng 7 2023

thanks

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MN

Minh Ngọc

8 tháng 2 2022

Tính nhanhA = \(\dfrac{1}{2}\) + \(^{\dfrac{1}{3}}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{63}\)Mình đang cần gấp , mong mn giúp mình với...

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#Toán lớp 6

2

NN

Như Nguyệt

8 tháng 2 2022

133/99

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VT

Vũ Trọng Hiếu

8 tháng 2 2022

quy đòng r tính nha ra \(\dfrac{199}{33}\)

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CT

Cao Tùng Lâm

26 tháng 1 2022

A = \(\dfrac{-19}{9}\times\dfrac{1}{2}-\dfrac{4}{11}\times\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)\)

B = \(\left(-\dfrac{15}{6}\right)\div\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}\times\dfrac{-11}{2}\)

C = \(\dfrac{3}{4}\times\left(-8\right)-\dfrac{1}{3}\times\dfrac{-7}{2}-\dfrac{5}{18}\)

#Toán lớp 6

2

MH

Minh Hiếu

26 tháng 1 2022

\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)

\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)

\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)

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NL

Nguyễn Lê Phước Thịnh

26 tháng 1 2022

\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)

\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)

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NT

Nguyễn Thanh Vân

29 tháng 1 2018

Giải các PT sau:

1) \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)

2) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{15}\)

Giúp mình với!!!

#Toán lớp 8

1

MP

Mysterious Person

6 tháng 2 2018

1) điều kiện xác định : \(x\notin\left\{-1;-3;-5;-7\right\}\)

ta có : \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\) \(\Leftrightarrow\dfrac{\left(x+5\right)\left(x+7\right)+\left(x+1\right)\left(x+7\right)+\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{x^2+12x+35+x^2+8x+7+x^2+4x+3}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{3x^2+24x+45}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow9\left(3x^2+24x+45\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x^2+8x+15\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27=\left(x+1\right)\left(x+7\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow27=x^2+8x+7\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-10\)

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MP

Mysterious Person

6 tháng 2 2018

câu còn lại lm tương tự nha bn

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VC

Vương Chí Hiếu

23 tháng 4 2023

1)\(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

2)\(\dfrac{ }{\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}}\)

3)\(\dfrac{ }{\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}}\)

4)\(\dfrac{ }{\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}}\)

#Toán lớp 6

1

H9

HT.Phong (9A5)

CTVHS

23 tháng 4 2023

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

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

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

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

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

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

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

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

\(=-\dfrac{35}{77}\)

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

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Tài nguyên

Ứng dụng mobile

\(\dfrac{1}{2}\) + \(\dfrac{1}{12}\)	2 + \(\dfrac{1}{6}\)	\(\dfrac{1}{20}\)	1 - \(\dfrac{1}{9}\)
\(\dfrac{1}{15}\) + \(\dfrac{2}{15}\)	\(\dfrac{1}{2}\) + \(\dfrac{2}{3}\)	\(\dfrac{7}{12}\)	\(\dfrac{4}{12}\)
\(\dfrac{9}{14}\)+ \(\dfrac{1}{14}\)	1 + \(\dfrac{1}{6}\)	\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)	\(\dfrac{1}{3}\) - \(\dfrac{2}{9}\)
\(\dfrac{3}{2}\) + \(\dfrac{2}{3}\)	\(\dfrac{1}{5}\)	1 - \(\dfrac{8}{9}\)
\(\dfrac{5}{7}\)	1 - \(\dfrac{2}{3}\)	\(\dfrac{1}{3}\) + \(\dfrac{5}{9}\)

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