\(\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy ...
\(4x^2-12x=-9\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+3^2=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy ...
\(\left(x+8\right)^2=121\)
\(\Rightarrow\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy ...
a.(x-4)2 -36=0
⇔(x-4-6)(x-4+6)=0
⇔(x-10))(x+2)=0
✱x-10=0 => x=10
✱ x+2 =0 => x=-2
Vậy x=10 và x=-2
b) 4x2 -12 + 9 =0
⇔ (2x)2 -2.2x.3 + 32 = 0
⇔(2x-3)2 =0
⇔2x-3=0
⇔ x= \(\dfrac{3}{2}\)
c) (x+8)2 -121=0
⇔ (x+8)2 -112 =0
⇔ (x+8-11)(x+8+11) =0
⇔ (x-3) (x+19) =0
\(\begin{matrix}x-3=0\\x+19=0\end{matrix}\) ⇔ \(\begin{matrix}x=3\\x=-19\end{matrix}\)
a) (x-4)2 - 36 = 0
<-> (x-4)2 - 62 = 0
<-> (x-4-6).(x-4+6) = 0
<-> (x-10).(x+2) = 0
x-10=0 hoặc x+2=0
x=0+10 x=0-2
x=10 x=-2
Vậy x=10 và x=-2
c) (x+8)2 = 121
<-> (x+8)2 - 121 = 0
<-> (x+8)2 - 112 = 0
<-> (x+8-11).(x+8+11) = 0
<-> (x-3).(x+19) = 0
<-> x-3 = 0 hoặc x+19 = 0
<-> x=0+3 x=0-19
<->x=3 x=-19
Vậy x=3 và x=-19
