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a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017
a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017

\(\left(2x-8\right).\left(x+13\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x-8=0\\x+13=0\end{cases}\Rightarrow\hept{\begin{cases}2x=8\\x=-13\end{cases}\Rightarrow}\hept{\begin{cases}x=4\\x=-13\end{cases}}}\)
Vậy \(x\in\left\{4;-13\right\}\)
= 2x2+36x-8x-104
= 2x2 +28x -104
2x-8=0 hayx+13=0
x=0hay x=-13
vay x=0va 13
bn tách ra thế này nha :
+) 2x - 8 = 0
2x = 0+8
2x = 8
x = 8 : 2
x = ....
+) x + 13 = o
x = 0 -13
x = .....
vậy x = .....hoặc....
(2x-8)(x+13)=0
>>2x-8=0 hoặc x+13=0
2x =0+8 x = 0-13
2x = 8 x =-13
x = 8:2
x =4
Vậy x= 4 hoặc x= -13
Đúng thì k nhé