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\(4x^2+12x+9=\left(2x+3\right)^2\)
\(25x^2+10x+5=\left(5x+1\right)^2+4\)( bạn xem lại đề )
\(x^2+6x+9x^2=10x^2+6x=2x\left(5x+3\right)\)
\(25x^2+10xy+y^2=\left(5x+y\right)^2\)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1\)
\(=\left(2x+1\right)^2\)
\(1+12x+36x^2\)
\(=1+2.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
a) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)
b) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Leftrightarrow2x^2-3x+6x-9=0\)
\(\Leftrightarrow x\left(2x-3\right)+3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)
\(25x^2-10x+1-y^2\)
\(=\left(5x-1\right)^2-y^2\)
\(=\left(5x-y-1\right)\left(5x+y-1\right)\)
\(4y^2-4x^2-4y+1\)
\(=\left(2y-1\right)^2-\left(2x\right)^2\)
\(=\left(2y+2x-1\right)\left(2y-2x-1\right)\)
\(-y^2+6y-9+x^2\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x+y-3\right)\left(x-y+3\right)\)
=(5x-3)(5x+3)