Ta có: \(\left(x+3\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+3\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=\dfrac{2}{3}\\x+3=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{3}\\x=\dfrac{-11}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\dfrac{7}{3};-\dfrac{11}{3}\right\}\)
\(\left(x+3\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+3\right)^2=\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{4}{9}\)
TH1 : \(x+3=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{3}-3=\dfrac{2}{3}-\dfrac{9}{3}=-\dfrac{7}{3}\)
TH2 : \(x+3=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{9}{3}=-\dfrac{11}{3}\)
Giải:
(x+3)2-1/3=1/9
(x+3)2 =1/9+1/3
(x+3)2 =4/9
=>(x+3)2 =(2/3)2
=>x+3 =2/3
x =2/3-3
x =-7/3
Chúc bạn học tốt!
Làm tiếp:
+)TH2:
=>(x+3)2 = (-2/3)2
x+3 = -2/3
x =-2/3-3
x =-11/3
Đầy đủ rồi nhé!
(x+3)^2-1/3=1/9
(x+3)^2=1/9+1/3
(x+3)^2=4/9
(x+3)^2=(2/3)^2
⇒x+3=2/3
*TH1:x+3=2/3 *TH2:x+3=-2/3
x=2/3-3 x=-2/3-3 x=-7/3 x=-11/3