Question

[Lê Phương Anh]

lm giúp mik vs ak

Answer 1

[Nguyễn Lê Phước Thịnh]

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: \(1-36x^2=0\)

\(\Leftrightarrow\left(6x-1\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Answer 2

[Nguyễn Hoàng Minh]

\(7,=\left(0,5a+5b\right)\left(0,25a^2-2,5ab+25b^2\right)\\ 8,=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\\ 9,=\left(5+a-b\right)\left(25-5a+5b+a^2-2ab+b^2\right)\)

\(61,\\ 1,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ 2,\Leftrightarrow\left(1-6x\right)\left(1+6x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ 3,\Leftrightarrow\left(6-x\right)\left(6+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)