b) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=\left(2x+1\right)\left(3x-5\right)\\ \Rightarrow2x-1=3x-5\\ \Rightarrow-x=-4\\ \Rightarrow x=4\)
c) \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)
\(\Rightarrow x^2+2x+1=4x^2-8x+4\)
\(\Rightarrow-3x^2+10x-3=0\\ \Rightarrow\left(-3x^2+9x\right)+\left(x-3\right)=0\\ \Rightarrow-3x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(1-3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-\left(2x+1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy ...
c, \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)
\(\Leftrightarrow-3x^2+10x-3=0\)
\(\Leftrightarrow-3x^2+x+9x-3=0\)
\(\Leftrightarrow x\left(-3x+1\right)-3\left(-3x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ....
