Tìm x, biết: \(\left(3x-1\right)^{2013}=\left(3x-1\right)^{2015}\)
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8 tháng 9 2023
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
11 tháng 3 2019
a.\(\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)
\(x-\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)
\(x-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{2}{3}\)
11 tháng 3 2019
\(a.\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\frac{5}{6}-x=\frac{1}{6}\)
\(\Leftrightarrow\frac{5}{6}-\frac{1}{6}=x\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(b.||3x+2|-2x-5|=3x-\left(-1\right)^{2015}\)
\(\Leftrightarrow||3x+2|-2x-5|=3x+1\)
\(\Leftrightarrow\orbr{\begin{cases}|3x+2|-2x-5=3x+1\\|3x+2|-2x-5=-3x-1\end{cases}\Leftrightarrow\orbr{\begin{cases}|3x+2|=5x+6\left(n\right)\\|3x+2|=-\left(x-4\right)\left(l\right)\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=5x+6\\3x+2=-5x-6\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x=4\\8x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)
V...\(x=-1;x=-2\)
26 tháng 10 2016
đầu bài trên tớ làm luôn nhá !!!
a, / 3x+1/= 5-3
/ 3x+1/= 2
3x+1=2
x+1 = 2:3
x+1 = 2 phần 3
x= 2/3 -1
x= -1/3
20 tháng 9 2017
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 ..
LN
20 tháng 9 2017
\(\left(4x-5\right)\left(3x+2\right)=0\)
\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
Nên:
\(x+2017=0\Rightarrow x=-2017\)
JY
4
5 tháng 2 2017
(3x - 7)2015 = (3x - 7)2017
(3x - 7)2017 - (3x - 7)2015 = 0
(3x - 7)2017[(3x - 7)2 - 1] = 0
=> (3x - 7)2017 = 0 hoặc (3x - 7)2 = 1
=> 3x - 7 = 0 hoặc 3x - 7 = ± 1
=> x = 7/3 hoặc x = { 8/3 ; 2 }
Vậy x = { 2; 7/3; 8/3 }
5 tháng 2 2017
\(y\left(y^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y^2-1=0\end{cases}}\)
x=1/3 hoặc x=2/3 bạn nhé.
pt <=> \(\left(3x-1\right)^{2013}-\left(3x-1\right)^{2015}=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(\left(3x-1\right)^2-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(3x-2\right)3x=0\)
\(\orbr{\begin{cases}x=0\\3x-1=0,3x-2=0\end{cases}}\)
Vậy x=0, x=1/3,hoặc x=2/3