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\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x^2+3x+1-2x^2-x+3=0\)
=>2x=-4
hay x=-2
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(2x^2+2x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x^2+x+\dfrac{1}{4}=0\)
hay \(x=-\dfrac{1}{2}\)
Điều kiện: \(x\ne\pm1\)
\(\frac{2x+1}{\left(x-1\right)^2}-\frac{2x+3}{x^2-1}=0\)
\(\Rightarrow\frac{\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)
\(\Rightarrow\frac{2x^2+3x+1-\left(2x^2+x-3\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)
\(\Rightarrow\frac{2x+4}{\left(x-1\right)^2\left(x+1\right)}=0\Rightarrow2x+4=0\Rightarrow x=-2\)(thỏa mãn điều kiện)
Vậy x = -2
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
`(x-1/2)^2 +(2x-1)^2=0`
\(\Rightarrow\left(x^2-x+\dfrac{1}{4}\right)+\left(4x^2-4x+1\right)=0\\ \Rightarrow x^2-x+\dfrac{1}{4}+4x^2-4x+1=0\\ \Rightarrow5x^2-5x+\dfrac{5}{4}=0\\ \Rightarrow\dfrac{5}{4}\left(4x^2-4x+1\right)=0\\ \Rightarrow\dfrac{5}{4}\left(2x-1\right)^2=0\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow2x=0+1\\ \Rightarrow2x=1\\ \Rightarrow x=\dfrac{1}{2}\)
Tìm x biết (2x+1)^2 =0
(2x+1)^2 =0
=>2x+1=0
2x=-1
x=\(\frac{-1}{2}\)
vậy......