\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)

\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

Bài 2:

\(\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{x-2}{x\left(x+2\right)}\)

\(=\dfrac{x+x+2+x-2}{x\left(x+2\right)}=\dfrac{3x}{x\left(x+2\right)}=\dfrac{3}{x+2}\)

Để 3/x+2 là số nguyên thì \(x+2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{-1;-3;1;-5\right\}\)

\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)

Mình làm phần tiếp của bạn Phạm Tuấn Đạt nha .

b) \(\left(x+6\right)^2-2x\left(x+6\right)+\left(x-6\right)\left(x+6\right)\)

\(=x^2+12x+36-2x^2-12x+x^2-36\)

\(=0\)

2. 

\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=-3\)

\(4x^2-4x+1-\left(4x^2-12x+x-3\right)=-3\)

\(4x^2-4x+1-4x^2+12x-x+3+3=0\)

\(7x+7=0\)

\(x=-1\)

câu a=219

câu b = 5 nha có 2 trường hợp xảy ra

câu c tương tự câu b nha bn

a=219

Còn lại mình ko biết

giải hản ra bn ơi