Giải:
a) \(\left(x+8\right)^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-11\\x+8=11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-19\\x=3\end{matrix}\right.\)
Vậy ...
b) \(x^2+8x+16=0\)
\(\Leftrightarrow x^2+2.4.x+4^2=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
c) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.3+3^2=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a, (x+8)2=121
<=>\(\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy x\(\in\){3;-19}
b,x2+8x+16=0
<=>(x+4)2=0
<=> x+4=0
<=>x=-4
Vậy x=-4
c,4x2-12x=-9
<=> 4x2-12x+9=0
<=> (2x-3)2=0
<=> 2x-3=0
<=> 2x=3
<=> x=1,5
Vậy x=1,5
