a, \(\left(x+2\right)^2-25=0\)
\(\Leftrightarrow\left(x+2\right)^2=25\\ \Leftrightarrow x+2=\pm5\)
Xét 2 trường hợp:
\(\)*\(x+2=5\\ \Leftrightarrow x=3\)
*\(x+2=-5\\ \Leftrightarrow x=-7\)
Vậy......
\(b,x\left(9x-1\right)-\left(3x+5\right)\left(3x-5\right)=1\)
\(\Leftrightarrow9x^2-x-\left(9x^2-25\right)=1\\ \Leftrightarrow9x^2-x-9x^2+25=1\\ \Leftrightarrow-x+25=1\\ \Leftrightarrow-x=1-25\\ \Leftrightarrow-x=-24\\ \Leftrightarrow x=24\)
Vậy.........
\(a.\left(x+2\right)^2-25=0\Leftrightarrow\left(x+2\right)^2=25\)
\(\Leftrightarrow x+2=\sqrt{25}=\pm5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{x_1=-7;x_2=3\right\}\)
\(b.x\left(9x-1\right)-\left(3x+5\right)\left(3x-5\right)=1\)
\(\Leftrightarrow9x^2-x-\left(9x^2-25\right)=1\)
\(\Leftrightarrow-x+25=1\Leftrightarrow-x=-24\)
\(\Leftrightarrow x=24\)
