a: Đặt A=0
=>(x-1/3)(x+1/3)=0
=>x=1/3 hoặc x=-1/3
b: Đặt B=0
=>(x+1)2=0
=>x=-1
\(a,x^2-\dfrac{1}{9}=0\\ x^2=\dfrac{1}{9}\\\)
x= \(\dfrac{1}{3};x=-\dfrac{1}{3}\)
\(b,x^2+2x+1=0\\ \left(x+1\right)^2=0\\ x+1=0\\ x=-1\)
`A=x^2-1/9`
Đặt `A=0`
`x^2-1/9=0`
`x^2=1/9`
\(x=\sqrt{\dfrac{1}{9}}\)
`=> x = {-1/3; 1/3}`
______________________________
`B = x^2 + 2x + 1`
Đặt `B=0`
`x^2+2x+1=0`
`(x+1)^2=0`
`=> x+1=0`
`x=0-1=-1`
Vậy `x=-1`