//\(2x\)/\(+1\)/\(3^{2015}+3^0=3^{2016}+1\)
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10 tháng 5 2018
\(||2x|+1|.3^{2015}+3^0=3^{2016}+1\)
\(\Leftrightarrow||2x|+1|.3^{2015}+1=3^{2016}+1\)
\(\Leftrightarrow||2x|+1|.3^{2015}=3^{2016}\)
\(\Leftrightarrow||2x|+1|=3\)
\(\Leftrightarrow\orbr{\begin{cases}|2x|+1=3\\|2x|+1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}|2x|=2\left(\text{chọn}\right)\\|2x|=-4\left(\text{loại}\right)\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
\(\text{Vậy}:S=\left\{-1;1\right\}\)
11 tháng 5 2018
\(||2x|+1|3^{2015}+3^0=3^{2016}+1\)
\(\Rightarrow||2x|+1|3^{2015}+1=3^{2016}+1\)
\(\Rightarrow||2x|+1|3^{2015}=3^{2016}+1-1\)
\(\Rightarrow||2x|+1|3^{2015}=3^{2016}\)
\(\Rightarrow||2x|+1|=3^{2016}\div3^{2015}\)
\(\Rightarrow||2x|+1|=3\)
\(\Rightarrow\orbr{\begin{cases}|2x|+1=3\\|2x|+1=-3\end{cases}\Rightarrow\orbr{\begin{cases}|2x|=3-1\\|2x|=-3-1\end{cases}\Rightarrow}\orbr{\begin{cases}|2x|=2\\|2x|=-4\end{cases}}}\)
\(\Rightarrow|2x|=2\)
\(\Rightarrow\orbr{\begin{cases}2x=2\\2x=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=2\div2\\x=-2\div2\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
20 tháng 5 2022
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
20 tháng 5 2022
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
20 tháng 4 2018
Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
20 tháng 4 2018
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~