K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HH
3
NN
5
F
11 tháng 6 2018
Trả lời :
\(x+3\frac{1}{4}+x=24\frac{1}{4}\)
\(x+\frac{13}{4}+x=\frac{97}{4}\)
\(x\text{ x 2 }+\frac{13}{4}=\frac{97}{4}\)
\(x\text{ x 2 }=\frac{97}{4}-\frac{13}{4}\)
\(x\text{ x 2 }=\frac{84}{4}=21\)
\(x=21:2\)
\(x=10,5\)
11 tháng 6 2018
\(x+3\frac{1}{4}+x=24\frac{1}{4}\)
\(\Rightarrow2x+\frac{13}{4}=\frac{97}{4}\)
\(\Rightarrow2x=21\)
\(\Rightarrow x=\frac{21}{2}\)
2 tháng 6 2020
\(A=(1\frac{1}{5}+\frac{17}{24}+\frac{29}{32}+\frac{1}{4}+\frac{7}{24}+\frac{3}{32})\div4\)
\(A=\left\{1\frac{1}{5}+\frac{1}{4}+(\frac{17}{24}+\frac{7}{24})+(\frac{29}{32}+\frac{3}{32})\right\}\div4\)
\(A=\left\{\frac{29}{20}+1+1\right\}\div4\)
\(A=\frac{69}{20}\div4\)
\(A=\frac{69}{80}=0,8625\)
2 tháng 6 2020
a=(6/5+17/24+29/32+1/4+7/24+3/32):4
a=[6/5+(17/24+7/24)+(29/32+3/32+1/4)]:4
a=[6/5+1+1+1/4):4
a=[24/20+20/20+20/20+5/20]:4
a=[44/20+25/20]:4
a=69/20:4
a=69/80
a=
TP
2
NX
27 tháng 8 2023
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)
\(D=\dfrac{1}{3}+\dfrac{2}{3}\)
\(D=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 8 2023
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)
D = \(\dfrac{3}{3}\)
D = 1
NM
3
TH
Thầy Hùng Olm
Manager
VIP
19 tháng 1 2023
=\(\dfrac{1}{18}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{33}+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{38}\)
=\(\dfrac{1}{18}-\dfrac{1}{38}=\dfrac{19-9}{38x9}=\dfrac{10}{342}=\dfrac{5}{171}\)
19 tháng 1 2023
\(\dfrac{5}{18\times23}+\dfrac{1}{23\times24}+\dfrac{7}{24\times31}+\dfrac{2}{31\times33}+\dfrac{4}{33\times37}+\dfrac{1}{37\tímes38}\)
\(\dfrac{23-18}{18\times23}+\dfrac{24-23}{23\times24}+\dfrac{31-24}{24\times31}+\dfrac{33-31}{31\times33}+\dfrac{37-33}{33\times37}+\dfrac{38-37}{37\times38}\)
\(\dfrac{23}{18\times23}-\dfrac{18}{18\times23}+\dfrac{24}{23\times24}-\dfrac{23}{23\times24}+\dfrac{31}{24\times31}-\dfrac{24}{24\times31}+\dfrac{33}{31\times33}-\dfrac{31}{31\times33}+\dfrac{37}{33\times37}-\dfrac{33}{33\times37}+\dfrac{38}{37\times38}-\dfrac{37}{37\times38}\)
\(\dfrac{1}{18}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{33}+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{38}\)
\(\dfrac{1}{18}-\dfrac{38}=\dfrac{5}{171}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 6 2023
\(\dfrac{1}{2}\) của 24\(\dfrac{1}{2}\) kg là : (24\(\dfrac{1}{2}\))\(\times\) \(\dfrac{1}{2}\) = \(\dfrac{49}{4}\)
d. \(\dfrac{49}{4}\)
NH
9 tháng 3 2019
\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(=\frac{1}{2}(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12})\)
\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{10}-\frac{1}{12})\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{12}\right)=\frac{1}{2}\cdot\frac{5}{12}=\frac{5}{24}\)
23 tháng 9 2021
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}=\dfrac{2}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{48}\\ =\dfrac{2}{3}-\dfrac{1}{48}=\dfrac{31}{48}\)
27 tháng 5 2015
a)\(\frac{1978x1979+1980x21+1958}{1980x1979-1978x1979+100}\)
\(=\frac{1978x1979+\left(1+1979\right)x21+1958}{\left(1980-1978\right)x1979+100}\)
\(=\frac{1978x1979+1979x21+21+1958}{2x1979+2x50}\)
\(=\frac{1979x\left(1978+21\right)+\left(21+1958\right)}{2x\left(1979+50\right)}\)
\(=\frac{1979x1999+1979x1}{2x2029}\)
\(=\frac{1979x2000}{2x2029}\)
\(=\frac{1979000}{2029}\)
đáp án: \(\frac{1}{4}\)
\(\frac{2}{4}\)- \(\frac{1}{4}\)=\(\frac{1}{4}\)