Bài 3:
1: Ta có: \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
3: Ta có: \(\left(x-5\right)^2-x\left(2x+1\right)=25\)
\(\Leftrightarrow x^2-10x+25-2x^2-x-25=0\)
\(\Leftrightarrow-x^2-11x=0\)
\(\Leftrightarrow-x\left(x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-11\end{matrix}\right.\)
4: Ta có: \(\left(3x-7\right)^2=x^2+8x+16\)
\(\Leftrightarrow\left(3x-7\right)^2-\left(x+4\right)^2=0\)
\(\Leftrightarrow\left(3x-7-x-4\right)\left(3x-7+x+4\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
2)=25-30x+9x^2
=5^2 - 2.5.(3x) + (3x)^2
=(5-3x)^2
