1: Ta có: \(x^2-16=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
2: Ta có: \(1-36x^2=0\)
\(\Leftrightarrow\left(6x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(7,=\left(0,5a+5b\right)\left(0,25a^2-2,5ab+25b^2\right)\\ 8,=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\\ 9,=\left(5+a-b\right)\left(25-5a+5b+a^2-2ab+b^2\right)\)
\(61,\\ 1,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ 2,\Leftrightarrow\left(1-6x\right)\left(1+6x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ 3,\Leftrightarrow\left(6-x\right)\left(6+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Câu 61:
1. x2 - 16 = 0
<=> x2 - 42 = 0
<=> (x - 4)(x + 4) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
b. 1 - 36x2 = 0
<=> 12 - (6x)2 = 0
<=> (1 - 6x)(1 + 6x) = 0
<=> \(\left[{}\begin{matrix}1-6x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
3. 36 - x2 = 0
<=> 62 - x2 = 0
<=> (6 - x)(6 + x) = 0
<=> \(\left[{}\begin{matrix}6-x=0\\6+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
