Giải:

\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Leftrightarrow4x^2+4x-1^2-\left(4x^2-12x+x-3\right)=3\)

\(\Leftrightarrow4x^2+4x-1-4x^2+12x-x+3=3\)

\(\Leftrightarrow15x+2=3\)

\(\Leftrightarrow15x=3-2\)

\(\Leftrightarrow15x=1\)

\(\Leftrightarrow x=\dfrac{1}{15}\)

Vậy \(x=\dfrac{1}{15}\).

Chúc bạn học tốt!!!

tìm x biết

x2-x+1/4=0

\(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

tìm x biết

a) x2(x+1)+2x(x+1)=0

Ta có:\(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Rightarrow\left(x^2+2x\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+2x=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{0;-1;-2\right\}\)