1. Tính: a.$\dfrac{1}{1.2} \dfrac{1}{2.3} ... \dfrac{1}{9.10}$ b.$\dfrac{1}{1.2} \dfrac{1}{2.3} \dfrac{1}{3.4} ... \dfrac{1}{99.100}$ 2. Tìm x , biết: a. \(\dfrac{2}{5} \dfrac{4}{5}x-\dfrac{7}...

MN

Mai Ngọc Trâm

10 tháng 6 2017

1. Tính: a.$\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}$ b.$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$ 2. Tìm x , biết: a. $\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}$ b. $\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}$ bài này ko được coppy trên...

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#Toán lớp 6

9

MS

Mashiro Shiina

10 tháng 6 2017

1)Tính

a)$\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}$

=$1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}$

$=1-\dfrac{1}{10}$

$=\dfrac{9}{10}$

b)$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}$

$=1-\dfrac{1}{100}$

$=\dfrac{99}{100}$

2) tìm x

$a$) $\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}$$=\dfrac{9}{5}$

$\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}$

$\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}$

$\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}$

$\dfrac{4}{5}x=0$

$x=0:\dfrac{4}{5}$

$x=0$

b)$\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}$

$\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}$

$\dfrac{2}{5}x=\dfrac{31}{10}$

$x=\dfrac{31}{10}:\dfrac{2}{5}$

$x=\dfrac{31}{4}$

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TX

Thịnh Xuân Vũ

10 tháng 6 2017

1. Tính:

a. $\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}$

= $\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}$

= $\dfrac{1}{1}-\dfrac{1}{10}$

= $\dfrac{10}{10}-\dfrac{1}{10}$

= $\dfrac{9}{10}$

b. $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$

= $\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

= $\dfrac{1}{1}-\dfrac{1}{100}$

= $\dfrac{100}{100}-\dfrac{1}{100}$

= $\dfrac{99}{100}$

2. Tìm x, biết:

a. $\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}$

$\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}$

$\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}$

$\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}$

$\dfrac{4}{5}x=\dfrac{14}{5}$

$x=\dfrac{14}{5}:\dfrac{4}{5}$

$x=\dfrac{14}{5}.\dfrac{5}{4}$

$x=14.\dfrac{1}{4}$

$x=\dfrac{14}{4}$

Vậy $x=\dfrac{14}{4}$

b. $\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}$

$\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}$

$\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}$

$\dfrac{2}{5}x=\dfrac{62}{20}$

$\dfrac{2}{5}x=\dfrac{31}{10}$

$x=\dfrac{31}{10}:\dfrac{2}{5}$

$x=\dfrac{31}{10}.\dfrac{5}{2}$

$x=\dfrac{31}{2}.\dfrac{2}{2}$

$x=\dfrac{31}{2}.1$

$x=\dfrac{31}{2}$

Vậy $x=\dfrac{31}{2}$

bài này mk tự làm ko sao chép trên mạng

nếu thấy đúng thì tick đúng cho mk nha

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T

Tiến

21 tháng 7 2023 - olm

Tìm x,biết: ($\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + ........ + $\dfrac{1}{8.9}$ + $\dfrac{1}{9.10}$) . 100 - [ $\dfrac{5}{2}$ : ($x+\dfrac{206}{100}$) ] : $\dfrac{1}{2}$ = 89 (Dấu . trong bài là dấu nhân...

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#Toán lớp 5

1

ND

Nguyễn Đức Trí

21 tháng 7 2023

$\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1$

$\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2$

$\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}$

$\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}$

$\Rightarrow x=-\dfrac{81}{100}$

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NA

Ngọc Anh

5 tháng 5 2022

Bài 1 : Tìm 2 số biết hiệu của chúng bằng 5 và 50% số lớn = 1 nửa số bé.
Bài 2 : tính giá trị biểu thức: A = $\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}$
Bài 3 : Tìm x
a , $\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}$

#Toán lớp 6

1

TC

TV Cuber

5 tháng 5 2022

bài 2:

$A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)$

$A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)$

$A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}$

bài 3:

$=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}$

$=>x=3$

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CV

Cherry Vương

10 tháng 3 2023

10.4. Tính tổnga) $\dfrac{1}{1}$ - $\dfrac{1}{2}$b) $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$c) $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ +...........$\dfrac{1}{99.100}$d) $\dfrac{3}{1.2}$ + $\dfrac{3}{2.3}$ +.........$\dfrac{1}{99.100}$giúp...

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#Toán lớp 6

1

NH

Ngô Hải Nam

10 tháng 3 2023

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

$\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}$

d)

$\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}$ đề phải như thế này chứ nhỉ?

$=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}$

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TN

Trần Ngọc Linh

14 tháng 8 2017

$a.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

#Toán lớp 6

2

EJ

Eren Jeager

14 tháng 8 2017

a, $\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178$

$\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178$

$\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178$

$90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178$

$\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)$ $=-88$

$x+\dfrac{206}{100}=\dfrac{-5}{176}$

$x=\dfrac{-5}{176}-\dfrac{206}{100}$

$x=\dfrac{-9198}{4400}$

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JP

Jenny Phạm

14 tháng 8 2017

a) $\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89$

$\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89$

$\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1$

$\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}$

$x+\dfrac{206}{100}=5$

$x=5-\dfrac{206}{100}$

$x=\dfrac{147}{50}$

Vậy $x=\dfrac{147}{50}$

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LN

Lê Ngọc Anh

20 tháng 3 2022

Bài 1: Thực hiện các phép tính:

d) 3,15+2,4=5,55

e) $\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}$

f) 1,25.3,6+3,6.8,75

h) B= $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$

#Toán lớp 6

2

VD

Vô danh

20 tháng 3 2022

d, `3,15+2,4=5,55`

e, $\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}$

f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`

$h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}$

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SM

Sơn Mai Thanh Hoàng

20 tháng 3 2022

$e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}$

$f3.6\times\left(1.25+8.75\right)=3.6\times10=36$

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NL

Nguyễn Lê Phương Linh

16 tháng 9 2023 - olm

$x-\dfrac{1}{1.2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-...-\dfrac{1}{98.99}=\dfrac{1}{100}+\dfrac{1}{99.100}$

#Toán lớp 7

2

AH

Akai Haruma

Giáo viên

16 tháng 9 2023

Lời giải:

$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$

$=1$

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KT

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』

16 tháng 9 2023

`# \text {DNamNgV}`

$x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}$

$\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}$

$\Rightarrow x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)=\dfrac{1}{99}$

$\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}$

$\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}$

$\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}$

$\Rightarrow x=\dfrac{99}{99}$

$\Rightarrow x=1$

Vậy, `x = 1.`

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NL

nguyễn lê bảo trâm

8 tháng 1 2021

1) Tính $\dfrac{7^4.3-7^3}{7^4.6-7^3.2}$ ; $\dfrac{10^3+5.10^2+5}{6^3+3.6^2+3^2}$ ; $E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}$ 2) Tìm x biết$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}$ ; $3^{x+1}+3^{x+3}=810$ MN ƠI ! GIÚP MIK VS > ....

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

8 tháng 1 2021

Bài 1:

a) Ta có: $\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}$

$=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}$

$=\dfrac{1}{2}$

c) Ta có: $E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}$

$\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}$

$\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)$

$\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}$

$\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}$

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NL

nguyễn lê bảo trâm

9 tháng 1 2021

thanks

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KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

11 tháng 2 2022

Chứng tỏ rằng:

a, $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}< 1$

b, $\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1$

#Toán lớp 6

1

TT

Trần Tuấn Hoàng

11 tháng 2 2022

a. $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$

$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=1-\dfrac{1}{100}< 1$.

b. Có: $\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}$.

$\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}< 1$

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B

BùiNgọcTháiHà16052010

26 tháng 4 2022 - olm

$\dfrac{x}{200}$= $\dfrac{1^2}{1.2}$ . $\dfrac{2^2}{2.3}$ . $\dfrac{3^2}{3.4}$ . .... .$\dfrac{99^2}{99.100}$

#Toán lớp 6

2

K

Kimchon

26 tháng 4 2022

bạn hãy rút gọn vế phải: x/200=1/2.2/3.3/4......98/99.99/100

Rồi sẽ có cái phương trình:x/200=1/100

từ đó suy ra:x/200=2/200 =>x=2

:)))))

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YN

Yen Nhi

27 tháng 4 2022

$\dfrac{x}{200}=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}$

$\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}$

$\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{100}$

$\Leftrightarrow x=2$

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