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5300.x = 340.(x+\(\frac{1}{15}\))
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/x+4/15/-/-3,75/=-/-2,15/
/x+4/15/-15/4=-\(\frac{43}{20}\)
x+4/15=\(\frac{43}{20}+\frac{15}{4}\)
x+4/15=59/10
x=59/10-4/15
x=\(5\frac{19}{30}\)
\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)
\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)
vậy x=-20 và y=-2
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Giải
Ta có:\(\frac{x}{5}+1=\frac{1}{y-1}\)
\(\Rightarrow\frac{x}{5}+\frac{5}{5}=\frac{1}{y-1}\)
\(\Rightarrow\frac{x+5}{5}=\frac{1}{y-1}\)
\(\Leftrightarrow\left(x+5\right).\left(y-1\right)=5\)
Vì \(x;y\in Z\)
\(\Rightarrow x+5;y-1\in Z\)
\(\Rightarrow x+5;y-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Ta lập bảng:
x + 5 | -5 | -1 | 1 | 5 |
y - 1 | -1 | -5 | 5 | 1 |
x | -10 | -6 | -4 | 0 |
y | 0 | -4 | 6 | 2 |
Vậy có 4 cặp ( x ; y ) cần tìm.
~~~~~~~~ *** ~~~~~
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}=\frac{x+3}{2008}\)
\(\Leftrightarrow\frac{x+4}{2007}=\frac{x+1}{2010}\)
\(\Leftrightarrow\left(x+4\right)2010=\left(x+1\right)2007\)
\(\Leftrightarrow2010x+8040=2007x+2007\)
\(\Leftrightarrow2010x-2007x=2007-8040\)
\(\Leftrightarrow3x=-6033\)
\(\Leftrightarrow x=-2011\)
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\frac{x+3}{2008}\)
=>\(\left(\frac{x\text{+4}}{2007}+1\right)+\left(\frac{x+8}{2003}+1\right)=\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+3}{2008}+1\right)\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}=\frac{x+2011}{2010}+\frac{x+2011}{2008}\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}-\frac{x+2011}{2010}-\frac{x+2011}{2008}=0\)
=>\(x+2011\left(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\ne0\)
=> x+2011=0
=>x=-2011
Vậy x = -2011
\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)
Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)
=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018
=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0
=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0
=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )
=> x=-2017
Vậy x=-2017
k mk nha
\(\dfrac{x}{6}=\dfrac{2x}{12}=\dfrac{y}{7}\)
\(\Rightarrow\dfrac{2x}{12}=\dfrac{y}{7}=\dfrac{2x-y}{12-7}=\dfrac{15}{5}=3\)
\(\Rightarrow x=3\cdot6=18\)
\(\Rightarrow y=3\cdot7=21\)
15 - ( 13 + x) = x- ( 23 - 17 )
15-13-x=x-23+17
2x=(-15)+(-13)-23+17
2x=(-28)-23+17
2x=(-51)+17
2x=(-34)
x=(-34):2
x=(-17)
15 - ( 13 + x ) = x - ( 23 - 17 )
15 - 13 - x = x - 6
-x - x = -6 + ( -2 )
-x + ( -x ) = -8
2.(-x) = -8
-x = -8 : 2
-x = -4
=> x = 4
Vậy x = 4
Chúc bạn học giỏi
\(5300x=340\left(x+\frac{1}{15}\right)\)
\(\Leftrightarrow5300x=340x+\frac{68}{3}\)
\(\Leftrightarrow5300x-340x=\frac{68}{3}\)
\(\Leftrightarrow4960x=\frac{68}{3}\)
\(\Leftrightarrow x=4960:\frac{68}{3}\)
\(\Leftrightarrow x=....\)(tự tính)
5300x=340x +\(\frac{68}{3}\)
4960x=\(\frac{68}{3}\)
x=\(\frac{34}{7440}\)