A= \(\left(\dfrac{1}{2}-1\right)\)\(\left(\dfrac{1}{3}-1\right)\).........\(\left(\dfrac{1}{10}-1\right)\). So sánh A với \(\dfrac{-1}{9}\)B= \(\left(\dfrac{1}{4}-1\right)\)\(\left(\dfrac{1}{9}-1\righ...

1

NL

Nguyễn Lê Phước Thịnh

31 tháng 12 2023

a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)

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

9<10

=>1/9>1/10

=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)

=>\(A>-\dfrac{1}{9}\)

b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

20<21

=>\(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

DT

Diễm Thuý Huỳnh Thị

19 tháng 10 2023

cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}-1\right).....\left(\dfrac{1}{10}-1\right)\). so sánh a với \(-\dfrac{1}{9}\)

1

H9

HT.Phong (9A5)

19 tháng 10 2023

Ta có:

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)

\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)

\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)

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

Mà: \(10>9\)

\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)

\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)

\(\Rightarrow A>-\dfrac{1}{9}\)

TT

Tri thức Việt

17 tháng 5 2022

Cho biểu thức \(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)

Hãy so sánh A với \(\dfrac{11}{19}\)

1

VL

Vui lòng để tên hiển thị

17 tháng 5 2022

`A = 3/4 xx 8/9 xx ... xx 99/100`

`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`

`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`

`= 1/10 xx 11`

`= 11/10`.

Ta có: `11/10 > 1`

`11/19 < 1`.

`=> A > 11/19`.

Y

ỵyjfdfj

23 tháng 9 2021

\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)....\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)

So sánh B với \(\dfrac{11}{21}\)

2

LL

Lấp La Lấp Lánh

23 tháng 9 2021

\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)

Bài 1. a, Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\) So sánh A với \(\dfrac{-1}{9}\) Bài 2. Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)....\left(\dfrac{1}{2008}-1\right)\left(\dfrac{1}{2009}-1\right)\) B = \(\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)....\left(-1\dfrac{1}{2007}\right)\left(-1\dfrac{1}{2008}\right)\) Tính A . B...

NH

Nguyễn Hoàng Minh

23 tháng 9 2021

\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)

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B

BK13

2 tháng 10 2017

3

N

ngonhuminh

2 tháng 10 2017

bai 1

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)

\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)

\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)

GT

Giang Thủy Tiên

2 tháng 10 2017

Violympic toán 7

a)\(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\) b) \(3-\left(17-x\right)=-12\)c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\) d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\) ...

MN

Miru nèe

22 tháng 6 2021

4

T

꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...

22 tháng 6 2021

Giải:

a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\)

\(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{11}\)

\(2x=\dfrac{16}{11}-\dfrac{7}{2}\)

\(2x=\dfrac{-45}{22}\)

\(x=\dfrac{-45}{22}:2\)

\(x=\dfrac{-45}{44}\)

b) \(3-\left(17-x\right)=-12\)

\(3-17+x=-12\)

\(x=-12-3+17\)

\(x=2\)

c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)

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

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

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

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

d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\)

\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\)

\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)

Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\)

e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)

\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\)

\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\)

\(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\)

\(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\)

\(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\)

\(-0,6x=\dfrac{-7}{18}\)

\(x=\dfrac{-7}{18}:-0.6\)

\(x=\dfrac{35}{54}\)

f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\)

\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\)

\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\)

\(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\)

\(\dfrac{3}{5}.x=\dfrac{13}{9}\)

\(x=\dfrac{13}{9}:\dfrac{3}{5}\)

\(x=\dfrac{65}{27}\)

Chúc bạn học tốt!

K

Kậu...chủ...nhỏ...!!!

22 tháng 6 2021

f)câu khó nhất

=>3x-1=0 và -1/2x+5=0

=>x=1/3 và x=10

VN

Võ Ngọc Phương

VIP

20 tháng 8 2023

Cho \(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)

So sánh \(A\) với \(\dfrac{-11}{21}\)

Giải chi tiết dùm mik nha. Thankss

4

ND

Nguyễn Đức Trí

VIP

20 tháng 8 2023

\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)

\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)

mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)

\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)

Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)

\(\Rightarrow A>-\dfrac{11}{21}\)

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WT

when the imposter is sus

20 tháng 8 2023

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)

\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)

\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)

Dễ thấy A có 9 thừa số, suy ra

\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)

Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)

Vậy \(A< \dfrac{-11}{21}\)

PD

Phạm Đăng Khoa

17 tháng 8 2021

Cho A=\(\left(\dfrac{1}{2^2}-1\right)\)\(\left(\dfrac{1}{3^2}-1\right)\)\(\left(\dfrac{1}{4^2}-1\right)\)...\(\left(\dfrac{1}{2013^2}-1\right)\)\(\left(\dfrac{1}{2014^2}-1\right)\) và B= \(-\dfrac{1}{2}\)

Hãy so sánh A và B

1

TQ

Trương Quang Khánh

17 tháng 8 2021

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B

NA

Người Âm

14 tháng 8 2021

\(a,\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2\)

\(b,\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3\)

c,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{4}\right)^4\)

\(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(\dfrac{-3}{2}\right)^3\)

1

NL

Nguyễn Lê Phước Thịnh

14 tháng 8 2021

a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)

b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)

c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)

d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)

Rút gọn các biểu thứcM = \(\sqrt{\left(3a-1\right)^2}+2a-3\) với a \(\ge\dfrac{1}{3}\)N = \(\sqrt{\left(4-a\right)^2}-a+5\) với a > 4I = \(\sqrt{\left(3-2a\right)^2}+2-7\) với a < \(\dfrac{3}{2}\)K = \(\dfrac{a^2-9}{4}\sqrt{\dfrac{4}{\left(a-2\right)^2}}\) với a <...

QE

Quynh Existn't

28 tháng 6 2021

#Toán lớp 9

2

Y

Yeutoanhoc

28 tháng 6 2021

`M=sqrt{(3a-1)^2}+2a-3`

`=|3a-1|+2a-3`

`=3a-1+2a-3(do \ a>=1/3)`

`=5a-4`

`N=sqrt{(4-a)^2}-a+5`

`=|4-a|-a+5`

`=a-4-a+5(do \ a>4)`

`=1`

`I=sqrt{(3-2a)^2}+2-7`

`=|3-2a|-5`

`=3-2a-5(do \ a<3/2)`

`=-2-2a`

`K=(a^2-9)/4*sqrt{4/(a-2)^2}`

`=(a^2-9)/4*|2/(a-2)|`

`=(a^2-9)/(2|a-2|)`

Nếu `3>a>2=>|a-2|=a-2`

`=>K=(a^2-9)/(2(a-2))`

Nếu `a<2=>|a-2|=2-a`

`=>K=(a^2-9)/(2(2-a))`

Đúng(3)

NN

Nguyễn Ngọc Lộc

28 tháng 6 2021

\(M=\left|3a-1\right|+2a-3\)

Mà \(a-\dfrac{1}{3}\ge0\)

\(\Rightarrow M=3a-1+2a-3=5a-4\)

\(N=\left|4-a\right|-a+5\)

Mà \(4-a< 0\)

\(\Rightarrow N=a-4-a+5=1\)

\(I=\left|3-2a\right|-5\)

Mà \(a-\dfrac{3}{2}< 0\)

\(\Rightarrow I=3-2a-5=-2a-2\)

K, Ta có : \(a-3< 0\)

\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)