\(A=\dfrac{1}{10} \dfrac{1}{11} \dfrac{1}{12} ... \dfrac{1}{99} \dfrac{1}{100}>1\) \(\dfrac{2n 5}{n 2}\) \(S=\dfrac{1}{20} \dfrac{1}{21} \dfrac{1}{22} L \dfrac{1}{29}\) \(A=\dfrac{7}{4}.\left(\dfra...

2

TQ

Trần Quảng Hà

24 tháng 4 2017

cho minh xin yeu cau de bai

PN

Phạm Nguyễn nguyên Anh

26 tháng 4 2017

trả hiểu yêu cầu đề bài là j cả

Tính nhanh a, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + \(\dfrac{1}{143}\) b, A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) c, H =\(\dfrac{4047991-2010x2009}{4050000-2011x2009}\) d, T = \(\dfrac{2009x20010+2000}{2011x2010-2020}\) e, P = \(\dfrac{7589-80,5x69,3}{7485,05-79x69,3}\) f, B = 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700 Giúp mình...

TD

Trần Đoàn Thanh Hiền

23 tháng 6 2023

#Toán lớp 5

1

VT

Vi Thị Vinh

25 tháng 6 2023

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

TN

Trần Nguyễn Lam My

18 tháng 3 2023

\(A=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{65}+...+\dfrac{1}{99}=\dfrac{16}{x}\)
\(B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{x}{16}\)\(C=\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{9}\right)\times\left(1-\dfrac{1}{16}\right)\times...\times\left(1-\dfrac{1}{100}\right)=\dfrac{22}{x}\)
m.ng giúp em với

#Toán lớp 5

1

NL

Nguyễn Lê Phước Thịnh

18 tháng 3 2023

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

H

huongff2k3

20 tháng 7 2021

tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

7

NV

Ngo VIP PRO

20 tháng 7 2021

trẻ trâu lửa chùa

hahahaha

H

huongff2k3

20 tháng 7 2021

địt mẹ mày mày bảo cái j đấy

Tính một cách hợp lý: a\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)\)) \(x:\dfrac{99}{100}:\dfrac{98}{99}:...:\dfrac{2}{3}:\dfrac{1}{2}\) b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\) c) \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\) d)...

DP

Đoàn Phương Linh

13 tháng 2 2018

1

R

Ruby

30 tháng 3 2018

b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)

Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé

DP

Đoàn Phương Linh

30 tháng 3 2018

cảm ơn bạn

\(\text{Bài 4. Chứng tỏ rằng:}\)\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17} 2\)\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}...

E

ᴵᴬᴹɴɢᴜʏễɴ●ɴɢọᴄ●ʜùɴɢッ

11 tháng 4 2021

2

NA

Ngoc Anh Thai

Giáo viên

11 tháng 4 2021

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

NA

Ngoc Anh Thai

Giáo viên

11 tháng 4 2021

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

Giải phương trình\(a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)\(b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)\(c,3\left(x-1\right)+3=5x\)\(d,\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)\(e,\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)\(f,\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)Em mới học về pt nên chưa quen lắm mọi người giúp e với ạ !Nguyễn Việt Lâm Quản lý ...

KH

Kamato Heiji

23 tháng 1 2021

#Toán lớp 8

3

NL

Nguyễn Lê Phước Thịnh

23 tháng 1 2021

a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-9=10x+85\)

\(\Leftrightarrow3x-10x=85+9\)

\(\Leftrightarrow-7x=94\)

hay \(x=-\dfrac{94}{7}\)

Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)

b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)

\(\Leftrightarrow6x-4-60=9-6x-42\)

\(\Leftrightarrow6x-64=-6x-33\)

\(\Leftrightarrow6x+6x=-33+64\)

\(\Leftrightarrow12x=31\)

hay \(x=\dfrac{31}{12}\)

Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)

c) Ta có: \(3\left(x-1\right)+3=5x\)

\(\Leftrightarrow3x-3+3=5x\)

\(\Leftrightarrow3x-5x=0\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: S={0}

d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)

\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)

\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)

mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)

nên x+101=0

hay x=-101

Vậy: S={-101}

👁💧👄💧👁

23 tháng 1 2021

a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)

Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt

b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)

Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt

c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt

d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)

Vậy x = -101 là nghiệm của pt

e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

Bài 1 : Thực hiện phép tính a/ \(\dfrac{7}{6}\) - \(\dfrac{13}{12}\) + \(\dfrac{3}{4}\)b/ 1 \(\dfrac{1}{2}\) . \(\dfrac{-4}{5}\) + \(\dfrac{3}{10}\)c/ \(\dfrac{25}{9}\) . \(\dfrac{3}{10}\) + ( \(\dfrac{-5}{3}\) )\(^2\) . \(\dfrac{7}{10}\) + | \(\dfrac{-25}{3}\) | Bài 2 : Tìm x , biếta/ x - \(\dfrac{5}{6}\) = \(\dfrac{1}{4}\)b/ \(\dfrac{26}{x}\) = \(\dfrac{-13}{-15}\)( Cần gấp...

C

Cihce

27 tháng 9 2021