Không biết nãy bị lỗi ở đâu, mình gửi lại:<

a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

Ta có: \(\dfrac{\left(x+3\right)\left(x-3\right)}{3}+2=x\left(1-x\right)\)

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

\(\Leftrightarrow x^2-9+6=3x-3x^2\)

\(\Leftrightarrow x^2-3-3x+3x^2=0\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)

Bài 1: 

\(P=2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)

\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)

\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)

\(=2x^2+4xy-7y^2\)

\(\left(x^2-6x\right)^2-2\left(x-3\right)^2=81\)

\(\Leftrightarrow\left(x^2-6x\right)^2-2\left(x^2-6x+9\right)=81\)

Đặt \(x^2-6x=t\), khi đó pt mang dạng:

\(t^2-2\left(t+9\right)=81\)\(\Leftrightarrow t^2-2t-18=81\)

\(\Leftrightarrow t^2-2t-99=0\Leftrightarrow t^2+9t-11t-99=0\)

\(\Leftrightarrow t\left(t+9\right)-11\left(t+9\right)=0\Leftrightarrow\left(t+9\right)\left(t-11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+9=0\\t-11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-6x+9=0\\x^2-6x-11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-6x+9=0\\x^2-2.x.3+9-20=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\\left(x-3\right)^2=20\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\sqrt{20}+3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\sqrt{5}+3\end{cases}}\)

Vậy tập nghiệm của pt là \(S=\left\{3;2\sqrt{5}+3\right\}.\)