Giải phương trình:
a)
\(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\\ \Leftrightarrow6x-18-\left(x-1-x-1\right)\left(x-1+x+1\right)=2x\\ \Leftrightarrow6x-18-4x=2x\)
\(\Leftrightarrow-18=0\) (vô lí)
=> Phương trình đã cho vô nghiệm
b)
\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1+2-x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
c)
\(4x^2-1=\left(2x+1\right)\left(3x-5\right)\\ \Leftrightarrow4x^2-1=6x^2-10x+3x-5\\ \Leftrightarrow4x^2-1=6x^2-7x-5\\ \Leftrightarrow4x^2-1-6x^2+7x+5=0\\ \Leftrightarrow-2x^2+7x+4=0\\ \Leftrightarrow2x^2-7x-4=0\\ \Leftrightarrow2x^2+x-8x-4=0\\ \Leftrightarrow x\left(2x+1\right)-4\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=4\end{matrix}\right.\)
d)
\(2x^2-x=3-6x\\ \Leftrightarrow2x^2-x-3+6x=0\\ \Leftrightarrow2x^2+5x-3=0\\ \Leftrightarrow2x^2+6x-x-3=0\\ \Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
e)
\(x\left(2x-7\right)-4x+14=0\\ \Leftrightarrow2x^2-7x-4x+14=0\\ \Leftrightarrow2x^2-11x+14=0\\ \Leftrightarrow2x^2-4x-7x+14=0\\ \Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
f)
\(\left(2x-5\right)^2-\left(x+3\right)^2=0\\ \Leftrightarrow\left(2x-5-x-3\right)\left(2x-5+x+3\right)=0\\ \Leftrightarrow\left(x-8\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=\frac{2}{3}\end{matrix}\right.\)
