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a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
Ta có : \(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}=4\)
\(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}-4=0\)
\(\dfrac{x-3}{2015}-1+\dfrac{x-4}{2014}-1+\dfrac{x-5}{2013}-1+\dfrac{x-6}{2012}-1=0\)
\(\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}+\dfrac{x-2018}{2013}+\dfrac{x-2018}{2012}=0\)
\(\left(x-2018\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}>0\)
=> x - 2018 = 0
x = 0 + 2018
x = 2018
Vậy x = 2018
\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)
<=> \(\frac{x+2014}{2}+\frac{2\left(x+2014\right)}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)
<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}=\frac{x+2014}{5}+\frac{x+2014}{6}\)
<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)
<=> \(\left(x+2014\right)\left(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\ne0\)
=> x + 2014 = 0 <=> x = -2014
Bài làm :
\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)
\(\Rightarrow\frac{x+2014}{2}+\frac{2x+4028}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)
\(\Rightarrow\frac{x+2014}{2}+\frac{2.\left(x+2014\right)}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)
\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Rightarrow x+2014=0:\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)\)
\(\Rightarrow x+2014=0\)
\(\Rightarrow x=-2014\)
Vậy x = - 2014 .
Học tốt nhé
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
Bạn tham khảo lời giải tại đây:
https://olm.vn/hoi-dap/detail/262254938778.html
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)
\(\left(\dfrac{x+1}{2015}+1\right)+\left(\dfrac{x+2}{2014}+1\right)+\left(\dfrac{x+3}{2013}+1\right)=0\)
\(\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)
\(\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow x+2016=0\Rightarrow x=-2016\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}=-3\)
\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1+\dfrac{x+3}{2013}+1=0\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}=0\)
\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow x+2016=0\Rightarrow x=-2016\)
Chúc bạn học tốt!!!
a/ \(\left(4x-5\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ............
b/ \(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x+2}{2015}+1\right)=\left(\dfrac{x+3}{2014}+1\right)+\left(\dfrac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Leftrightarrow x+2017\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
\(\left(4x-5\right)\left(3x+2\right)=0\)
\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
Nên:
\(x+2017=0\Rightarrow x=-2017\)
\(\dfrac{x+2014}{2}+\dfrac{2\left(x+2014\right)}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)
\(\left(x+2014\right)\left(\dfrac{1}{2}+\dfrac{2}{7}\right)=\left(x+2014\right)\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(\left(x+2014\right)\dfrac{11}{14}=\left(x+2014\right)\dfrac{11}{30}\)
Dấu ''=''↔x=-2014