11+5.(x-1)2=56
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H
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
15 tháng 1
\(D=\dfrac{6-1}{1.6}+\dfrac{11-6}{6x11}+\dfrac{16-11}{11x16}+...+\dfrac{56-51}{51.56}=\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{51}-\dfrac{1}{56}=1-\dfrac{1}{56}=\dfrac{55}{56}\)
DQ
1
26 tháng 7 2023
\(=\dfrac{2}{5}\cdot\dfrac{5}{8}\cdot...\cdot\dfrac{53}{56}=\dfrac{2}{56}=\dfrac{1}{28}\)
C
12 tháng 2 2023
`#qlv`
`(11)/(25) . (y + y5 - (23)/(55)) + 3/(14) . (56)/(25) = 1`
`=> (11)/(25) . (6y - (23)/(55)) + (12)/(25) = 1`
`=> (11)/(25) . (6y - (24)/(55)) = 1 - (12)/(25)`
`=> (11)/(25) (6y - (23)/(55)) = (13)/(25)`
`=> 6y - (23)/(55) = (13)/(25) : (11)/(25)`
`=> 6y - (23)/(55) = (13)/(11)`
`=> 6y = (13)/(11) + (23)/(55)`
`=> 6y = 8/5`
`=> y = 8/5 : 6`
`=> y = 8/5 . 1/6`
`=> y = 4/(15)`
Vậy `y = 4/(15)`
Y
12 tháng 2 2023
\(\dfrac{11}{25}\times\left(y+y\times5-\dfrac{23}{55}\right)+\dfrac{3}{14}\times\dfrac{56}{25}=1\)
\(\Rightarrow\dfrac{11}{25}\times\left(10\times y-\dfrac{23}{55}\right)+\dfrac{12}{25}-1=0\)
\(\Rightarrow\dfrac{22}{5}\times y-\dfrac{23}{125}-\dfrac{13}{25}=0\)
\(\Rightarrow\dfrac{22}{5}\times y=\dfrac{88}{125}\)
\(\Rightarrow y=\dfrac{88}{125}:\dfrac{22}{5}\)
\(\Rightarrow y=\dfrac{4}{25}\)
27 tháng 4 2017
1.
a, 2/3=12/x
12/18=12/x
=>x=18
Vậy x=18
b, 14/56=1/x
1/4=1/x
=>x=4
Vậy x=4
c, 24/36=x/12
2/3=x/12
8/12=x/12
=>x=8
Vậy x =8
2
38/11+4/17-5/11-3/17
=(38/11-5/11)+(4/17-3/17)
=33/11+1/17
=3+1/17
=51/17+1/17
=52/17
3.
2/7 x =2/3
x=2/3:2/7
x=2/3x7/2
x=7/3
Vậy x=7/3
2/5:x=1/3
x=2/5:1/3
x=2/5x3
x=6/5
Vậy x= 6/5
K cho mình nha bạn !!!!!!!!!!!!^^
HM
0
A
21 tháng 4 2019
\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot56\)
\(\Rightarrow\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot56\)
\(\Rightarrow\left(\frac{6+11}{33}\right)\cdot x=\left(\frac{8-7}{56}\right)\cdot56\)
\(\Rightarrow\frac{17}{33}\cdot x=\frac{1}{56}\cdot56\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=\frac{33}{17}\)
Vậy x = 33/17
21 tháng 4 2019
\(\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{1}{7}-\frac{1}{8}\right]\cdot56\)
\(\Rightarrow\left[\frac{2\cdot3}{33}+\frac{1\cdot11}{33}\right]\cdot x=\left[\frac{8}{56}-\frac{7}{56}\right]\cdot56\)
\(\Rightarrow\left[\frac{6}{33}+\frac{11}{33}\right]\cdot x=\frac{1}{56}\cdot56\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=1\cdot\frac{33}{17}=\frac{33}{17}\)
VT
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 9 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{11}{12}\) + \(\dfrac{19}{20}\) + \(\dfrac{29}{30}\) + \(\dfrac{41}{42}\) + \(\dfrac{55}{56}\)
A = (1 - \(\dfrac{1}{2}\)) + ( 1 - \(\dfrac{1}{6}\)) + (1 - \(\dfrac{1}{12}\)) + (1 - \(\dfrac{1}{20}\)) +(1-\(\dfrac{1}{30}\))+(1-\(\dfrac{1}{42}\))+(1-\(\dfrac{1}{56}\))
A = (1 + 1+1 + 1 + 1+1+1)- (\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\))
A = 7 - (\(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{8}\))
A = 7 - \(\dfrac{7}{8}\)
A = \(\dfrac{49}{8}\)
\(11+5.\left(x-1\right)^2=56\)
\(=>5.\left(x-1\right)^2=56-11\)
\(=>5.\left(x-1\right)^2=45\)
\(=>\left(x-1\right)^2=45:5\)
\(=>\left(x-1\right)^2=9\)
\(=>\left(x-1\right)^2=3^2\) hoặc \(\left(x-1\right)^2=\left(-3\right)^2\)
\(=>x-1=3\) hoặc \(x-1=-3\)
\(=>x=3+1\) hoặc \(x=\left(-3\right)+1\)
\(=>x=4\) hoặc \(x=-2\)
Vậy \(x=4\) hoặc \(x=-2\)
11 + 5.( \(x\) - 1)2 = 56
5.(\(x\) - 1)2 = 56 - 11
5.(\(x\) - 1)2 = 45
\(\left(x-1\right)\)2 = 45 : 5
(\(x\) - 1)2 = 9
\(\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3+1\\x=3+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2; 4}