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AH
Akai Haruma
Giáo viên
15 tháng 7 2023

Lời giải:

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$

$\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$

$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$
$1-\frac{1}{4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$

$\frac{1}{x\times (x+1)}=\frac{-1}{4}$ (đây là số âm lớp 4 chưa học). Bạn xem lại đề.

30 tháng 3 2022

tham khảo:

30 tháng 3 2022

99/100

3 tháng 5 2023

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

3 tháng 5 2023

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)\(\times\) \(\dfrac{8}{7}\): 2

\(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

\(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

\(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)\(\times\) ( 1 - \(\dfrac{1}{3}\)\(\times\) ( 1 - \(\dfrac{1}{4}\)\(\times\) ( 1 - \(\dfrac{1}{5}\))

\(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

\(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

\(\dfrac{1}{5}\)

11 tháng 7 2023

a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)

\(=1-\dfrac{1}{256}\)

\(=\dfrac{255}{256}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{13}{14}\)

c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)

\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)

\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)

\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=\dfrac{1}{15}-\dfrac{1}{90}\)

\(=\dfrac{6}{90}-\dfrac{1}{90}\)

\(=\dfrac{5}{90}=\dfrac{1}{18}\)

 

11 tháng 7 2023

tớ đang cần gấp

 

25 tháng 2 2020

1/2+4/3*1/6

=1/2+4/18

=13/18

25 tháng 2 2020

1/2 + 4/3 = 3/6 + 8/6 = 11/6

4/5 + 3/4 : 2/3 = 4/5 + 3/4 * 3/2 = 4/5 + 9/8 = 32/40 + 45/40 = 77/40

7/2 - 4/3 * 1/6 = 7/2 - 2/9 = 63/18 - 4/18 = 59/18

4/5 * 4/7 : 2/3 = 16/35 : 2/3 = 16/35 * 3/2 = 24/35

1/4 : 5/3 + 1/6 = 1/4 * 3/5 + 1/6 = 3/20 + 1/6 = 9/60 + 10/60 = 19/60

3 - 4/5 : 1/3 = 3 - 4/5 * 3/1 = 3 - 12/5 = 15/5 - 12/5 = 3/5

5/9 * 2/7 - 3/7 = 10/63 - 3/7 = 10/63 - 27/63 = -17/63  

29 tháng 3 2023

`1/2 xx 1/3 xx 1/4`

`= (1xx1xx1)/(2xx3xx4)`

`= 1/24`

__

`1/2 xx 1/3 : 1/4`

`= 1/2 xx 1/3 xx 4`

`= (1xx1xx4)/(2xx3)`

`= 4/6`

`=2/3`

__

`1/2 : 1/3 xx1/4`

`= 1/2 xx 3 xx 1/4`

`=(1xx3xx1)/(2xx4)`

`= 3/8`

__

`1/2 : 1/3 : 1/4`

`= 1/2 xx 3xx4`

`= 12/2`

`=6`

29 tháng 3 2023

`1/2xx1/3xx1/4`

`=1/24`

 

`1/2xx1/3:1/4`

`=1/6xx4`

`=4/6=2/3`

 

`1/2:1/3xx1/4`

`=1/2xx3xx1/4`

`=3/2xx1/4`

`=3/8`

 

`1/2:1/3:1/4`

`=1/2xx3xx4`

`=6`

14 tháng 5 2023

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)

17 tháng 3 2022

bn ơi bn tìm trên google á

coá nhoa bn