Bài 3:
a: Ta có: \(x^2+4x-12=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-10x-24=0\)
\(\Leftrightarrow\left(x-12\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
c: Ta có: \(4x^2-4x-15=0\)
\(\Leftrightarrow\left(2x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
bài 4:
a: Ta có: \(A=x^2+6x+10\)
\(=x^2+6x+9+1\)
\(=\left(x+3\right)^2+1>0\forall x\)
b: Ta có: \(B=25x^2-10x+2\)
\(=25x^2-10x+1+1\)
\(=\left(5x-1\right)^2+1>0\forall x\)
c: Ta có: \(C=x^2-3x+4\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}>0\forall x\)
