1/.$\dfrac{1}{1}.\dfrac{1}{2} \dfrac{1}{2}.\dfrac{1}{3} \dfrac{1}{3}.\dfrac{1}{4} \dfrac{1}{4}.\dfrac{1}{5}$ 2/.$\dfrac{1}{2} \dfrac{1}{6} \dfrac{1}{12} ... \dfrac{1}{10100}$ 3/.A = \(\dfrac{2}{...

5

NN

nguyen ngoc huyen

3 tháng 3 2017

2/ = $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ +......+$\dfrac{1}{100.101}$

= 1-$\dfrac{1}{2}$ +$\dfrac{1}{2}$ -$\dfrac{1}{3}$+.........+$\dfrac{1}{100}$-$\dfrac{1}{101}$

=1-$\dfrac{1}{101}$=...........

mk làm vậy thôi nha

thông cảm

LB

Lê Bảo Linh

2 tháng 3 2017

=$\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}$=$1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}$

=1-$\dfrac{1}{5}=\dfrac{4}{5}$

tương tự

Tính (theo mẫu).Mẫu: $2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}$a) $1+\dfrac{4}{9}$ b) $5+\dfrac{1}{2}$ c) $3-\dfrac{5}{6}$ ...

ML

Minh Lệ

24 tháng 8 2023

#Toán lớp 4

1

GL

Gia Linh

24 tháng 8 2023

a) $1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}$

b) $5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}$

c) $3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}$

d) $\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}$

LH

Lưu huỳnh ngọc

11 tháng 8 2021

1/ $\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}$

2/ $\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}$

3/ $\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}$

4/ $\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}$

5/ $\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0$

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

11 tháng 8 2021

1: Ta có: $\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}$

$\Leftrightarrow5x+1-2\left(x-2\right)=4$

$\Leftrightarrow5x+1-2x+4=4$

$\Leftrightarrow3x=-1$

hay $x=-\dfrac{1}{3}$

2: Ta có: $\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}$

$\Leftrightarrow9x+27+12-36x=-2x+2$

$\Leftrightarrow-27x+2x=2-39$

hay $x=\dfrac{37}{25}$

3: Ta có: $\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}$

$\Leftrightarrow3x+6-10x=4-4x$

$\Leftrightarrow-7x+4x=4-6=-2$

hay $x=\dfrac{2}{3}$

NL

Nguyễn Lê Phước Thịnh

11 tháng 8 2021

4: Ta có: $\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}$

$\Leftrightarrow5x-15-x-1=2x-4$

$\Leftrightarrow4x-2x=-4+16=12$

hay x=6

5: Ta có: $\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0$

$\Leftrightarrow12x+3-9x+5+4x-8=0$

$\Leftrightarrow7x=0$

hay x=0

LH

Lưu huỳnh ngọc

11 tháng 8 2021

1/ $\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}$

2/ $\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}$

3/ $\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}$

4/ $\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}$

5/ $\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}$

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

11 tháng 8 2021

1: Ta có: $\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}$

$\Leftrightarrow5x+20+12x-28=7x+2$

$\Leftrightarrow17x-7x=2+8=10$

hay x=1

2: Ta có: $\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}$

$\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}$

$\Leftrightarrow6x+4-12x=-3x+3$

$\Leftrightarrow-6x+3x=3-4$

hay $x=\dfrac{1}{3}$

3: Ta có: $\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}$

$\Leftrightarrow4x-12-x-2=6x-3$

$\Leftrightarrow3x-14-6x+3=0$

$\Leftrightarrow-3x=11$

hay $x=-\dfrac{11}{3}$

NL

Nguyễn Lê Phước Thịnh

11 tháng 8 2021

4: Ta có: $\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}$

$\Leftrightarrow3x-6-8x-12=x+6$

$\Leftrightarrow-5x-x=6+18$

hay x=-4

5: Ta có: $\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}$

$\Leftrightarrow6x-3+2x-6=-1$

$\Leftrightarrow8x=8$

hay x=1

tìm x bt a) $\dfrac{2}{3}$x - $\dfrac{7}{6}$ = $\dfrac{12}{7}$ - $\dfrac{1}{2}$ b) ( $1\dfrac{1}{2}$ + $\dfrac{5}{3}$ - $\dfrac{1}{6}$ ) : x...

TG

Trần Gia Hân

10 tháng 3 2023

1

AH

Akai Haruma

Giáo viên

10 tháng 3 2023

Lời giải:

a.

$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$

$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$

$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$

b.

$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$

$3:x=\frac{1}{4}$

$x=3: \frac{1}{4}=12$

$\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

Tính giá trị biểu thứca, $19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}$ b,$\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}$c, $\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}$ d, $\left[6+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{1}{2}\right|\right]:\dfrac{3}{12}$ ...

KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

27 tháng 1 2022

3

R

Rhider

27 tháng 1 2022

a) $=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}$

b) $\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}$

c) $=-\dfrac{80}{9}$

Đúng(3)

R

Rhider

27 tháng 1 2022

d) $=-\dfrac{1}{6}$

Tính (theo mẫu).Mẫu: $\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}$a) $\dfrac{3}{4}-\dfrac{1}{8}$ b) $\dfrac{2}{6}-\dfrac{5}{18}$ ...

ML

Minh Lệ

24 tháng 8 2023

#Toán lớp 4

1

GL

Gia Linh

24 tháng 8 2023

a) $\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}$

b) $\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}$

c) $\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}$

a) 0,25-$\dfrac{2}{3}$+1$\dfrac{1}{4}$b) $\dfrac{3^2}{2}$:$\dfrac{1}{4}$+$\dfrac{3}{4}$.2010c) {[($\dfrac{1}{25}$-0,6)2:$\dfrac{49}{125}$].$\dfrac{5}{6}$}-[($\dfrac{-1}{3}$)+$\dfrac{1}{2}$]d) (-$\dfrac{1}{2}$-$\dfrac{1^{ }}{3}$)2:[($\dfrac{-5}{36}$)-($\dfrac{-5}{36}$)0]Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like...

BM

bé mèo miu

1 tháng 10 2023

1

T

Toru

1 tháng 10 2023

a) $0,25-\dfrac{2}{3}+1\dfrac{1}{4}$

$=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}$

$=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}$

$=\dfrac{10}{12}$

$=\dfrac{5}{6}$

$---$

b) $\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010$

$=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}$

$=18+\dfrac{3015}{2}$

$=\dfrac{36}{2}+\dfrac{3015}{2}$

$=\dfrac{3051}{2}$

$---$

c) $\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]$

$=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]$

$=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}$

$=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}$

$=\dfrac{4}{6}-\dfrac{1}{6}$

$=\dfrac{3}{6}$

$=\dfrac{1}{2}$

$---$

d) $\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]$

$=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]$

$=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]$

$=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)$

$=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)$

$=-\dfrac{25}{41}$

#$Toru$

BM

bé mèo miu

1 tháng 10 2023

cảm ơn nhiều nha vừa kịp giờ lun