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\(\left|x-\frac{3}{5}\right|+2x-1=0\)
\(\Leftrightarrow2x-1=-\left|x-\frac{3}{5}\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=-x-\frac{3}{5}\\2x-1=-\left(-x\right)-\frac{3}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x+x=\frac{-3}{5}+1\\2x-x=\frac{-3}{5}+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=\frac{2}{5}\\x=\frac{2}{5}\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{15}\\x=\frac{2}{5}\end{cases}}\)
Vậy ..................
Chắc cách làm như thế :V
a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)
vậy phương trình vô nghiệm
a) (x-1)2 = 0
<=> x-1 = 0
<=> x = 1
b) (x-2)2 - 1 = 0
<=> (x-2)2 = 1
<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) (2x-1)3 = -8
<=> (2x-1)3 = -23
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(-\dfrac{1}{2}\)
d) (x+2)2 + 1 = 0
<=> (x+2)2 = -1
<=> x+2 = -1
<=> x = -3
1)
\(2^{x-1}=16\\ 2^{x-1}=2^4\\ \Rightarrow x-1=4\\ x=4+1\\ x=5\)
5)
\(\left(x-1\right)^2=25\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6)
\(\left|2x-1\right|=5\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
5) (x-1)2 = 25
(x-1)2 = 52
x-1 = 5
x = 5+1
x = 6
6) \(\left|2x-1\right|=5 \)
\(TH1:\) \(2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\)
\(TH2:2x-1=-5\)
\(\Leftrightarrow2x=-5+1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4:2\)
\(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2.
Tick nha!
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)
Vì GTTĐ luôn lớn hơn hoặc bằng 0, mà theo đề bài
=> 6 - 2x = 0 và x - 13 = 0
2x = 6 x = 13
x = 3
Vậy,................
|6-2x|+|x-13|=0
=>6-2x=0 hoặc x-13=0
6-2x=0 ; x-13=0
=>2x=6-0=6 =>x=13
=>x=6:2=3
vậy x thuộc {3;13}