\(A=\sqrt{9a^2-12a+4}-9a+1\)

\(A=\sqrt{9a^2-6a-6a+4}-9a+1\)

\(A=\sqrt{\left(9a^2-6a\right)-\left(6a-4\right)}-9a+1\)

\(A=\sqrt{3a.\left(3a-2\right)-2.\left(3a-2\right)}-9a+1\)

\(A=\sqrt{\left(3a-2\right)^2}-9a+1\)

\(A=3a-2-9a+1\)

\(A=-6a-1\)

\(A=-\left(6a+1\right)\)

Chúc bạn học tốt!!!

Cách làm :

Ta có : A =

<=> A = ( biến đổi)

Vậy A =

\(\sqrt{9a^2-12a+4}-9a+1\)

=\(\sqrt{\left(3a\right)^2-2.3a.2+2^2}-9a+1\)

=\(\sqrt{\left(3a-2\right)^2}-9a+1\)

=\(|3a-2|-9a+1\)

=\(3a-2-9a+1\)

=\(-6a-1\)

Thay \(a=\frac{1}{3}\)ta có:

\(-6.\frac{1}{3}-1\)

= \(-3\)

mình nhầm khúc này nha

=\(|3a-2|-9a+1\)

=\(2-3a-9a+1\)

=\(3-12a\)

Thay a=1/3. ta có

\(3-12.\frac{1}{3}\)

=\(3-4\)

=\(-1\)

ukm cảm ơn bạn 

\(=a\left(9a^2-12a+4\right)=a\left(3a-2\right)^2\)

\(9a^3-12a^2+4a=a\left(9a^2-12a+4\right)=a\left(3a-2\right)^2\)

\(9a^3-12a^2+4a\Leftrightarrow a\left[\left(3a\right)^2-2.3a.2+2^2\right]\) ⇔ \(a\left[\left(3a-2\right)^2\right]\)

\(9a^4-12a^2b^3+9b^6\)

\(=\left(3a^2\right)^2-2.3a^2.3b^3+\left(3a^3\right)^2\)

\(=\left(3a^2-3a^3\right)^2\)